This note includes very basic Topology, including point set topology and algebraic topology.
Might extend to differential topology in the future.
Continuous mappings \(f:\mathbb R^n \to \mathbb R^m\) is usually defined using metrics: \(\forall \epsilon > 0, \exists \delta > 0\), s.t. \(d(x,x_0) < \delta\) gives \(d(f(x),f(x_0)) < \epsilon\). But can we define continuity without any notion of metric?
The answer is yes. We can define continuity using only topological structures on sets.
Suppose we have a set \(X\). A topology \(\mathcal T\) is a set of the subsets of \(X\) with the following properties:
\(\varnothing, X\in \mathcal T\).
(Closed under arbitrary union) \(\forall i \in \mathcal I\), \(U_i \in \mathcal T\), then \(\cup_{i\in \mathcal I} U_i \in \mathcal T\).
(Closed under finite intersection) \(U_1,\dots, U_n\), then \(\cap_{i=1}^n U_i\in \mathcal T\).
Remark. A subset \(U \in \mathcal T\) is also called an open set. A subset \(U\in X\) is called a closed set if and only if \(X - U\) is open.
Some examples.
\(\mathcal T = \{\varnothing, X\}\), trivial topology.
Power set: \(\mathcal T = P(X) := \{\text{All subsets of } X\}\), dubbed discrete topology.
Let \(X\) be a set. A metric on \(X\) is a mapping \(X \times X \to \mathbb R_{\geq 0}\), subject to the following.
\(d(x,y) = 0\ \Leftrightarrow\ x = y\).
\(d(x,y) = d(y,x), \forall x,y \in X\).
(Triangular inequality) \(d(x,y) + d(y,z) \geq d(x,z), \forall x,y,z \in X\).
Let \(X\) be a set and \(d\) is a metric on \(X\). We define an open ball at \(x_0\) with radius \(\epsilon\) as \[B(x_0,\epsilon) := \{x\in X \ |\ d(x,x_0) < \epsilon\}.\]
Let \(X\) be a set and \(d\) is a metric on \(X\). Then for any subset \(U \in X\), \(U\) is open if and only if \[\forall x\in U, \exists \epsilon > 0, s.t. B(x,\epsilon) \subseteq U\]
Remark. This definition satisfies all the requirements of topological space. The proof is trivial class="definition" id="topology" data-attribution="Topology".
Moreover, all subset \(U\in \mathcal T\) can be written as the union of open balls. \[U = \cup_{x\in U} B(x,\epsilon).\]
Let \((X,\mathcal T)\) be a topological space. Suppose \(S\subset X\) is a subset of \(X\), then a subspace topology on \(S\) induced by \(\mathcal T\) is \[\mathcal T^\prime = \{U\in S\ |\ U = S\cap V, V \in \mathcal T\}.\]
Remark. Note that an open set \(U \subset S, U \in \mathcal T^\prime\) is not necessarily open at \((X,\mathcal T)\) (i.e., \(U\in \mathcal T\)). A simple example is \(S = [0,1]\subset \mathbb R\) and \(\mathcal T = \{(a,b)\ |\ a,b\in \mathbb R\}\), where \(U = [0,0.5)\in \mathcal T^\prime\) but \(U \not\in \mathcal T\).
Now we have defined topology on sets. It’s time to try defining continuity without using the notion of metric. Remember that in metric topology, we can define continuity as \(\forall \epsilon > 0, \exists \delta > 0\), \(|x - x_0|<\delta\) yields \(|f(x) - f(x_0)| <\epsilon\). This is in fact equivalent to \(f^{-1}\big(f(x_0) - \epsilon, f(x_0 + \epsilon)\big)\) is open in \(\mathbb R\).
Let \(X,Y\) be two topological space. We call a mapping \(f: X \to Y\) continuous, if \[\forall U \subseteq Y\ \text{that is open}, f^{-1}(U)\subseteq X\ \text{is open.}\]
Remark. The composition of continuous mappings is also continuous. Also observe that definition 1.6 reduces to the continuous mapping in Calculus when \(Y\) is a metric topology, in which case we can show \(\forall x\in X,\epsilon > 0, \exists U\in \mathcal T_X\ \text{with}\ x\in \mathcal T_X, U \subseteq B(f(x),\epsilon)\) is equivalent to definition 1.6.
Definition 1.6 also provides us a useful trick for defining a topology in space \(X\). Let \(\mathcal T_X^{f} := \{f^{-1}(U): U \in \mathcal T_Y\}\), then \(\mathcal T_X^{(f)}\) is a topology on \(f^{-1}(Y)\subseteq X\).
The definition of continuous mapping bridges across two topological space.
Let \(X\), \(Y\) be two topological space. A mapping \(f:X \to Y\) is a homeomorphism, denoted \(X \cong Y\), if
\(f\) is continuous
\(f\) is bijective.
\(f^{-1}\) is also continuous.
Remark. Homeomorphism is typically hard to disprove. It would take some effort to proof \(\mathbb R\not\cong \mathbb R^2\), and even more effort to show \(\mathbb R^2 \not\cong \mathbb R^3\).
Some examples.
(Embedding) Let \(f:X\to Y\) be continuous and injective. \(f\) is called an embedding from \(X\) to \(Y\), if \(f: X \to f(X)\) is homeomorphism.
\(f:\mathbb R \to \mathbb R^2\), \(f(x) = (ax,bx)\) is an embedding.
(Stereographic projection) \(S^2 = \{(x,y,z)\in \mathbb R^3\ |\ x^2 + y^2 + z^2 = 1\}\). Then \(S^2 \backslash \{(0,0,1)\} \cong \mathbb R^2\).
\(\varphi(x,y,z) = (\frac{2x}{1-z}, \frac{2y}{1-z} -1)\) and \(\varphi^{-1}(x,y-1) = (\frac{4x}{4+x^2 + y^2}, \frac{4y}{4+x^2+y^2}, \frac{-4+x^2 + y^2}{4+x^2 + y^2})\)
Next, we build a useful tool for proving continuous mappings. The following lemma allows us to check the continuity of a mapping by checking continuity in its segments.
Let \(X,Y\) be two topological spaces, \(\{X_i\ |\ i\in I\}\) be a family of subspaces of \(X\). Let \(\{f_i\ |\ i\in I\}\) be a family of continuous functions and \(f_i|_{X_i\cap X_j} = f_j|_{X_i\cap X_j}\) for \(\forall i,j\in I\). If
\(X_i\) are open sets in \(X\).
Or, \(X_i\) are closed and \(I\) is finite.
Then \(f:X\to Y\) with \(f|_{X_i} = f_i\) is continuous.
Let \(X\) and \(Y\) be two topological space. We want to find topology on \(X\times Y\). However, \(U\times V, U\in \mathcal T_X, V\in \mathcal T_Y\) does not naturally form a topology, as it is not closed under union.
Let \((X,\mathcal T_X)\), \((Y,\mathcal T_Y)\) be two topological space. We define open boxes in \(X\times Y\) as \(U\times V\), where \(U\in\mathcal T_X\) and \(V\in \mathcal T_Y\).
Moreover, we call a subset \(W\subseteq X\times Y\) open, if it is a union of open boxes.
Remark. To check this is indeed a topology is trivial. The most crucial property is that intersecting two open boxes \((U_1\times V_1) \cap (U_2\times V_2) = (U_1\cap U_2)\times (V_1\cap V_2)\) is still an open box.
We have seen in previous examples that a topology space may be constructed from some basic ingredients, e.g., open balls or open boxes. In this section, we formalize this intuition by characterising open balls and open boxed as basis.
Let \(X\) be a set. \(\mathcal B\) be a family of subsets of \(X\). We call \(\mathcal B\) a basis for a topology if
(Covers \(X\)) \(\forall x\in \mathcal B, \exists U\in \mathcal B\), such that \(x\in U\).
(Intersection) If \(U,V\in \mathcal B\), there exists a collection of subsets \(\{W_i\in \mathcal B\ |\ i\in I\}\) such that \(U\cap V = \cup_{i\in I}W_i\).
Let \(X\) be a set. \(S \subseteq 2^X\) is a subbasis if \[\bigcup_{S_i\in S} S_i = X. \quad \text{(only need S cover X)}\]
Given a subbasis, how do we generate a basis? \[\text{Subbasis}\ S \longrightarrow \text{Basis}\ \mathcal B \longrightarrow \text{Topology}\ \mathcal T\]
Let \(\mathcal B_S = \{S_{i_1}\cap\dots \cap S_{i_m}\ |\ S_{i_1},\dots, S_{i_m} \in S\}\). Then \(\mathcal B_S\) is a basis.
Let \(\mathcal T_\mathcal B = \{\bigcup_{i\in \mathcal I} B_i\ |\ B_i \in \mathcal B\}\). Then \(\mathcal T_\mathcal B\) is a topology.
Examples.
Metric topology \((X,d)\), corresponding to basis open balls \(\{B(x,\epsilon)\ |\ x\in X, \epsilon > 0\}\).
Product topology \(X\times Y\), corresponding to basis open boxes \(\{U\times V\ |\ U\in X, V\in Y\}\).
\(S = \{(-\infty, a)\ |\ a\in \mathbb R\} \cup \{(b,\infty)\ |\ b\in \mathbb R\}\) is a subbasis in \(\mathbb R\). Its corresponding basis is \(\mathcal B_S = \{(a,b)\ |\ \forall a,b\in \mathbb R\}\) which is already a topology in \(\mathbb R\).
(Initial topology) Consider some continuous mappings \(f_i : X \to Y_i\). Let \(S = \{f_i^{-1}(U)\ |\ U\subset Y_i\ \text{open}\}\). Then \(\mathcal T_S\) is called an initial topology.
\(f: X \to Y\), \(S\) is a subbasis of the topology of \(Y\). Then \[f\ \text{is continuous} \Leftrightarrow f^{-1}(S_i) \in \mathcal T_X, \forall S_i \in S\]
A topological space \(X\) is called Hausdorff if for any pair of points \(x,y\in X\), there exists an open neighborhood \(U\) of \(x\) and an open neighborhood \(V\) of \(y\) such that \(U\cap V = \varnothing\).
Relation. A relation in \(X\) is a subset \(R\subset X\times X\). \((x,y)\in R\) is denoted \(x\sim y\).
An equivalent relation satisfies the following properties:
(reflexive) \(x \sim x\)
(symmetric) \(x\sim y \Rightarrow y\sim x\)
(transitive) \(x\sim y, y\sim z \Rightarrow x\sim z\)
Consider two mappings \(p_1, p_2 :T \to X\). We can define \(p: T \to X\times X\), such that \(p(t) = (p_1(t),p_2(t))\). Then \(R_{p_1p_2} = p(T)\) is a relation generated by \(p_1\) and \(p_2\).
Let \(\sim\) be an equivalent relation. An equivalent class containing \(x\) is denoted by \[ [x]:= \{y\in X\ |\ x\sim y\}.\]
The quotient set in \(X\) of an equivalent relation \(\sim\) is denoted by \[X/\sim\ := \{[x]\ |\ x\in X\}.\]
\((X,\mathcal T_X)\) is a topology space. Let \(f:X \to Y\) be a surjective quotient. \(U\subset Y\) is open if \(f^{-1}(U)\) is open. The quotient topology is defined as \[\mathcal T_Y = \{U\subset Y \ |\ f^{-1}(U) \in \mathcal T_X\}\]
Remark.
\(\mathcal T_Y\) is the finest topology that makes \(f\) continuous.
For an equivalent relation \(\sim\), consider its projection map \(\pi: X \to X/ \sim\). The quotient topology is \[\mathcal T = \{U\subset X / \sim\ |\ \pi^{-1}(U)\in\mathcal T_X\}\]
Let \(X\) and \(Y\) be topological spaces. Suppose \(f:X \to Y\) is a continuous and surjective. If the quotient topology is homeomorphic to the topology on \(Y\), i.e., \((X/\sim_f, QT)\cong (Y,\mathcal T_Y)\), then \(f\) is called a quotient map.
The question is: how do we decide a map is a quotient map or not? The following propositions 2.5 and 2.6 are usually used jointly to prove a quotient map.
Let \(f:X\to Y\) be continuous and surjective. Let \(\{U_i\}_{i\in \mathcal I}\) be an open cover of \(Y\). We denote the restriction of \(f\) on \(U_i\) by \(f_i: f^{-1}(U_i) \to U_i\) with \(f_i = f|_{f^{-1}(U_i)}\).
Let \(f:X\to Y\) be continuous and surjective. If \(f\) has a section \(S:Y\to X\), i.e., \(S:Y\to X\) continuous and \(f\circ S = Id_Y\), then \(f\) is a quotient map.
Remark. The existence of a section \(S\) implies \(f\) is surjective.
Let \(f: X\to Y\) be a continuous, surjective map and \(f\) is open (or closed) map. Then \(f\) is a quotient map.
An open gluing datum consists of the following:
(What to glue) A family of topological spaces \(\{U_i\}_{i\in \mathcal I}\).
(Where to glue) Open subspaces \(U_{ij}\subset U_i\), we also set \(U_{ii} =U_i\).
(How to glue) homeomorphism \(f_{ij} : U_{ij} \to U_{ji}\)
satisfying the following
\(f_{ii} = Id_{U_i}\)
\(f_{ij}(U_{ij}\cap U_{ik})\subset U_{ji}\cap U_{jk}\)
(Cocycle condition) The homeomorphisms agree on triple intersections \[f_{ik}|_{U_{ij}\cap U_{ik}} = f_{jk}|_{U_{ji}\cap U_{jk}} \circ f_{ij}|_{U_{ij}\cap U_{ik}}\]
Remark. The cocycle condition guarantees the consistency when gluing \(> 2\) subsets together.
The disjoint union of a collection of sets \(\{U_i\}_{i\in \mathcal I}\): \[\bigsqcup_{i\in \mathcal I}U_i = \{(x,i)\ |\ i\in\mathcal I, x\in U_i\}\]
An open gluing \[ (U_i, U_{ij}, f_{ij}) \] is defined as \[U = \bigsqcup_{i\in \mathcal I}U_i.\] \[\sim: (x,i)\sim (y,i) \Leftrightarrow x\in U_{ij}, y\in U_{ji}, f_{ij}(x) = y.\]
Check that this is indeed an equivalence relation by definition. We call \((U/\sim, QT)\) the topological space glued from datum \((U_i,U_{ij}, f_{ij})\).
Let \(X\) be a topological space. An \(n\)-dimensional chart on \(X\) consists of an open subset \(U\subset X\) equipped with the subspace topology and a homeomorphism \(\phi: U \to V\), where \(V\) is an open subspace of \(\mathbb R^n\) with the standard topology.
Let \(M\) be a Hausdorff space with a countable topological basis (also called second countable). \(M\) is a topological manifold (of pure dimension \(n\)) if there exists a collection of (\(n\)- dimensional) charts \((U_i, \varphi_i)_{i\in I}\) such that \(M = \bigcup_{i\in I} U_i\).
A topological space \(X\) is connected if it is not the union of two disjoint non-empty open subsets.
Remark.
\(X\) is connected. \(\Leftrightarrow\) The subsets that are open and closed in \(X\) must be \(\varnothing\) and \(X\).
Let \(f: X \to Y\) be a continuous map of topological spaces. Assume \(X\) is connected, then \(f(X)\) is connected.
In order words, connectedness is a topological invariant.
Using this property, we can easily show \(\mathbb R \not\cong \mathbb R^2\). Since \(f(\mathbb R\backslash\{0\})\) is a connected subspace of \(\mathbb R^2\).
Let \(X\) be a topological space and \(X^\prime\subseteq X\) is a connected dense subset, then \(X\) is also connected.
Remark. A subset \(A\subseteq X\) is dense, if its closed envelop \(\overline A = X\), where \(\overline A = \bigcap_{i\in I} V_i\), \(V_i\) is closed and \(A\subseteq V_i\). An example: \(\overline {\mathbb Q} = \mathbb R\).
Suppose \(X\) is not connected, then there exists a nonempty proper subset \(U\subset X\) that is closed and open. Since \(X^\prime\) is dense, \(U\cap X^\prime \neq \varnothing\). Then \(U\cap X^\prime \subseteq X^\prime\) is closed and open in \(X^\prime\), which means \(U\cap X^\prime = X^\prime\). Thus, \(U = X^\prime = \overline X^\prime = X\), contradict.
If \(A\subseteq X\) and \(A\) is connected, then its closed envelop \(\overline A\) is also connected.
Let \(\Lambda\) be a topological space with discrete topology. A continuous map \(f:X \to \Lambda\) is locally constant if \[\forall x, \exists U\in\mathcal T, x\in U, \ \text{such that}\ f|_{U} \ \text{is constant}.\]
Let \((\Lambda, discrete)\) be a topological space with at least two elements. \(X\) is a topological space. Then \(X\) is connected if and only if any locally constant function \(f:X\to \Lambda\) is constant.
Using proposition 3.5, we can prove the following convenient proposition for connectedness.
Let \(\{A_i\}_{i\in I}\) be a collection of connected subspaces of \(X\). Let \(B\) be a connected subspace such that \(B\cap A_i \neq \varnothing\), then \[B\cup \bigcup_{i\in I} A_i\ \text{is connected.}\]
Define \(A = B\cup \bigcup_{i\in I}A_i\). Let \(f:A \to \Lambda\) be a locally constant function. We know \(f|_{A_i}\) and \(f_{B}\) are all constant function as \(A_i\) and \(B\) are connected. Thus, \(f|_A = f|_{U}\) is constant.
\((X,\mathcal T)\), if \(\forall x,y\in X\), there exists a connected subspace \(U\), \(x,y\in U\), then \(X\) is connected.
\(X,Y\) are connected \(\Rightarrow\) \(X\times Y\) is connected.
\(f:X\to Y\) is a surjective continuous map. \(Y\) has the quotient topology. Then \(X\) is connected \(\Rightarrow\) \(Y\) is connected.
Remark. Based on these corollaries above, we can show some basic spaces are connected.
\(\mathbb R^n\) is connected. (from corollary 3.8)
\(S^n\) is connected. Proof: for \(x,y\in S^n\), choose a \(z\in S^n\) and \(z\neq x,z\neq y\). Then \(\mathbb R^n / \{z\}\cong \mathbb R^n\) is connected.
\(\mathbb RP^n\) and \(\mathbb CP^n\) connected from corollary 3.9.
Let \(X\) be a topological space. A path \(\gamma\) is a continuous map \(\gamma: I \to X\). \(\gamma(0)\) and \(\gamma(1)\) are the start and the end of the path.
A topological space \(X\) is path connected if for any \(x,y\in X\), \(\exists\) a path \(\gamma\) with \(\gamma(0) = x\), \(\gamma(1) = y\).
\((X,\mathcal T)\). \(x,y\) are mutually connected if there is a connected subspace \(U\subset X\) and \(x,y\in U\).
Mutual connectedness is an equivalence relation. Its equivalence class are the connected components of \(X\).
Total number of connected components is a topological invariant: \(\pi_0(X)\).
Without any difficulty we can extend the mutual connectedness to mutual path connectedness. Note that mutual path connectedness is also an equivalence relation. Similarly we define path connected components.
Compactness seeks to generalize the property of a bounded and closed subset of Euclidean space.
Let \(X,\mathcal T\) be a topological space. \(C\) is compact if \(\forall\) open cover \(\{u_i\}_{i\in\mathcal I}\) of \(X\), there exists \(U_{i_1},\dots, U_{i_n}\in \{U_i\}_{i\in \mathcal I}\), such that \(X\subset \bigcup_{i=1}^n U_{i_j}\).
Remark. In other words, any open cover of \(X\) contains a finite open cover of \(X\).
Examples.
\((X,trivial)\) is compact. If \(|X| <\infty\), \((X,discrete)\) is compact.
\([0,1]\) is compact.
\((0,1)\) is not compact, because \(\{(\frac{1}{n},1)\}_{n=1}^\infty\) is an open cover, but no finite subset of it is a cover.
If \(f\) is continuous, \(X\) is compact, then \(f(X)\) is also compact.
\(X\) is a compact, closed space. \(Z\) is a closed subspace of \(X\). Then \(Z\) is also compact.
Remark. This proposition 4.3 guarantees that every bounded closed subset of \(\mathbb R\) is compact.
\(X\) is Hausdorff. \(Z\subset X\) is compact \(\Rightarrow\) \(Z\) is closed.
\(\mathcal U = \{U_i\}_{i\in I},\mathcal V = \{V_i\}_{i\in I}\) are open covers of \(X\). If \(\forall V_i,\exists U_k\), such that \(V_i \subset U_k\), then \(\mathcal V\) is called a refinement of \(\mathcal U\).
If \(X\), \(Y\) are compact topological spaces, then \(X\times Y\) is also compact.
Let \(\mathcal U\) be an open cover of \(X\times Y\), we can find a refinement of \(\mathcal V\) of \(\mathcal U\), where \(\mathcal V\) only contains the open boxes in \(X\times Y\). We can easily find a finite open cover in \(\mathcal V\) using compactness of \(X\) and \(Y\).
Let \(Z\) be a subspace of \(\mathbb R^n\). \(Z\) is compact if and only if \(Z\) is a bounded and closed subset.
If \(Z\) is compact, then since \(\mathbb R^n\) is Hausdorff, we know \(Z\) is closed according to proposition 4.4. Note that \(Z\subset \bigcup_{r>0}B(0,r)\), so \(Z \subset \bigcup_{r=r_1,\dots, r_n}B(0,r)\), which means \(Z\) is bounded.
The reverse direction is direct since \(Z\) is a closed subspace of \([-N,N]^n\) for sufficiently large \(N\).
In this section, we consider the set of all continuous map from \(X\) to \(Y\), dubbed \(\mathcal C[X;Y]\), and assign a topology on it using compactness.
Consider the set of all continuous map from \(X\) to \(Y\): \(\mathcal C[X;Y]\). Define a subbasis = \(\{\mathcal M_{K,U}\ |\ K\subset X\ \text{compact}, U\subset Y\ \text{open}\}\), where \[\mathcal M_{K,U} = \{f\in \mathcal C[X;Y]\ |\ f(K)\subset U\}.\] The topology generated by this subbasis is called a compact open topology.
Moreover, if \((Y,d)\) is a metric topological space, then we have
\((Y,d)\) metric space. A collection of sets \[B_K(f,\epsilon) = \{g\in \mathcal C[X;Y]\ |\ \sup_{x\in K} d(f(x),g(x)) < \epsilon\}.\] with \(K\subset X\) compact. Then \(\{B_K(f,\epsilon)\}\) is a basis of compact open topology.
Example.
The solution space for \(\ddot x + x = 0\): \(\mathcal M = \{x = a\cos t + b\sin t\ |\ (a,b)\in \mathbb R^2\}\) is homeomorphic to \(\mathbb R^2\).
\(Y\) can be embedded into \(\mathcal C[X;Y]\). Consider a bijective map \(c: Y \to \mathcal C[X;Y]\) such that \(y \mapsto f(X) = y\).
\(\mathcal C[X;Y]\) is Hausdorff if and only if \(Y\) is Hausdorff.
\(\mathcal C[X;\mathbb R]\) is path connected.
\(f: X \to \mathbb R\). Let \(c_f:[0,1]\to \mathcal C[X;\mathbb R]: c_f(t) = t\cdot f\).
Homeomorphism is good for sure, but usually too restricted to be held. Here we discuss a weaker condition, i.e., Homotopy.
Let \(f,g: X \to Y\) be two continuous maps. A homotopy between \(f\) and \(g\) is a continuous map \(\sigma: X\times [0,1]\to Y\) such that \(\forall x\in X\), \[\sigma(x,0) = f(x),\ \sigma(x,1) = g(x).\] \(f\) and \(g\) is then homotopic, denoted \(f\simeq g\).
Remark.
\(\sigma\) defines a path in \(\mathcal C[X;Y]\).
Be cautious that \(\sigma\) must be continuous on \(X \times [0,1]\). Only continuous for \(\sigma(\cdot, t)\) is not enough.
If \(f\) is homotopic to a constant function, it is called null-homotopic.
Homotopy \(f\simeq g\) is an equivalence relation.
Remark. We denote the quotient set \(\mathcal C[X;Y]/\simeq\) as \([X;Y]\).
Let \(f_1,f_2: X\to Y\) and \(g_1,g_2: Y\to Z\). If \(f_1\simeq f_2\) and \(g_1\simeq g_2\), then \[g_1\circ f_1 \simeq g_2\circ f_2.\]
We prove by \(g_1\circ f_2 \simeq g_1 \circ f_1 \simeq g_2 \circ f_2\).
The first relation can be proved by \(g_1 \circ \sigma_1\). The second can be shown by \(\sigma_2^\prime(x,t) = \sigma_2(f_2(x),t)\).
We consider the homotopic inverse. For \(f: X\to Y\) we require its homotopic inverse to satisfy: \(f\circ g \simeq Id_Y\) and \(g\circ f \simeq Id_X\).
Topological spaces \(X\) and \(Y\) are called homotopy equivalent if there exist continuous maps \(f:X\to Y\) and \(g:Y\to X\) such that \[f\circ g \simeq Id_Y,\ g\circ f \simeq Id_X.\]
Examples.
\(\mathbb R^n\simeq \{x\}\).
X is a contractible space if \(X\simeq \{x\}\). This is equivalent to the condition that \(Id_X\) is null-homotopic.
Example.
\(\mathbb R\backslash\{0\} \simeq \{p_1\}\cup \{p_2\}\).
\(\mathbb R^2 \backslash \{0\} \simeq S^1\).
To better describe the above observation, we define defomation retraction.
Let \(X\) be a topological space and \(A\) be a subspace of \(X\). A deformation retraction is a homotopy \[r: X\times [0,1] \to X\] such that \(r(x,0) = x\) and \(r(x,1) \in A\). Moreover, for any \(a\in A\) and \(t\in [0,1]\) we have \(r(a,t) = a\).
Let \(X\) be a topological space and \(\gamma:[0,1]\to X\) is a path on X. If \(\gamma(0) = \gamma(1)\) we call it a loop.
The inverse of a path \(\gamma\) is \(\gamma^- (t) = \gamma(1-t)\). The constant path \(I[x]\) is defined as \(\gamma(t) = x\) for some \(x\in X\).
Two paths with same initial and end points: \(\gamma_1,\gamma_2: [0,1]\to X\). \(\gamma_1\simeq_p \gamma_2\) if there exists a continuous map \(\sigma:[0,1]\times [0,1]\to X\) with \(\sigma(t,0) = \gamma_1(t)\) and \(\sigma(t,1) = \gamma_2(t)\). Moreover, we need \(\sigma(0,s) = \gamma_1(0) = \gamma_2(0)\) and \(\sigma(1,s) = \gamma_1(1) = \gamma_2(1)\).
Remark. Path homotopy is stricter than map homotopy as it fixes the initial and end points during deformation. To see their difference, consider \(\mathbb R\backslash \{0\}\).
Let \(\alpha, \beta\) be paths on a topological space \(X\). Suppose \(\alpha(1) = \beta(0)\). We define \[(\alpha *\beta)(s) = \begin{cases} \alpha(2s),\ s\in [0,\frac{1}{2}],\\ \beta(2s-1),\ s\in [\frac{1}{2},1]. \end{cases}\]
For a path \(\gamma\) on \(X\) we have \[\gamma * \gamma^- \simeq_p I[\gamma(0)]\]
Note that \(\alpha\simeq_p \alpha^\prime\) and \(\beta \simeq_p \beta^\prime\) yields \(\alpha*\beta \simeq_p \alpha^\prime * \beta^\prime\). We can define \[_p*[\beta]_p = [\alpha*\beta]_p\]
Suppose \(\alpha,\beta,\gamma\) are paths on \(X\). Then \[_p*([\beta]_p *[\gamma]_p) = ([\alpha]_p*[\beta]_p) * [\gamma]_p.\]
We now want to use define a group structure on this operation. However, not all paths on \(X\) can be multiplied as they might not have overlapped end points. To fix this issue, we consider loops with a fixed end points.
Let \(X\) be a topological space. Let \(x\in X\). We define \[\pi_1(X,x) = \{[\gamma]_p\ |\ \gamma\ \text{loops in}\ X,\ i.e., \gamma(0) = \gamma(1) = x\}.\] Then \((\pi_1(X,x), *)\) is a group.
Remark.
On \(\mathbb R^2\) all loops are path homotopic to the constant path. On \(\mathbb R^2\backslash\{0\}\), however, loops that wrap around \(0\) are not path homotopic to the constant path.
A fundamental group is not necessarily Abelian.
\(\pi_1((X,x),*)\) is a homotopic invariant.
Examples.
Let \(A\) be a band and \(B\) be a Mobius band. Then \(\pi_1(A,*) \cong \pi_1(B,*)\cong \pi_1(S^1,*)\cong (\mathbb Z,+)\).
If \(X\) is path connected, then \(\pi_1((X,x),*)\cong \pi_1((X,y),*)\) for any \(x,y\in X\).
Remark. Let \(X\) be a topological space, then the followings are equivalent
\(\pi_1(X,x)\) is trivial
\(\forall \gamma_1,\gamma_2\) paths with \(\gamma_1(0) = \gamma_2(0)=x\) and \(\gamma_1(1) = \gamma_2(1)\), we have \(\gamma_1\simeq_p\gamma_2\).
All loops at \(x\) \(\simeq_p\) constant loop at \(x\).
Let \((E,p)\) be a space over \(p\) with \(p:E \to B\) continuous surjective. \(p\) is a covering map and \((E,p)\) is a covering space over \(B\), if \(\forall b\in B\) there exists an open neighbor \(U\) of \(b\) such that \[p^{-1}(U) = \bigcup_{i\in I} U_i\] where \(U_i\) are disjoint open subsets of \(E\) and \(p|_{U_i} :U_i \to U\) is a homeomorphism.
Let \(p: E \to B\) be a covering map. Let \(c: [0,1]\to B\) be a path on \(B\). If \(\tilde b \in E, p(\tilde b) = c(0) = b\), then there exists a unique path in \(E\): \(\tilde c: [0,1]\to E\), such that \(\tilde c(0) = \tilde b\) and \(p\circ \tilde c = c\).
Let \(p: E \to B\) be a covering map. Let \(c_1,c_2:[0,1]\to B\) be two homotopic paths on \(B\). Suppose \(c_1(0) = c_2(0) = b\). Then \(\tilde c_1, \tilde c_2\) with \(\tilde c_1(0) = \tilde c_2(0) = \tilde b \in p^{-1}(b)\) are also homotopic.
In discussion of path-connected components, we note that \[\pi_0(X) = \{[x\to X]\ |\ x\in X\}\] can be viewed as the set of all homotopic equivalent classes from a single point to \(X\). Similarly, \[\pi_1(X) = \{[[0,1]\to X]\ |\ x\in X\}\] can be viewed as the set of all homotopic equivalent classes of loops in \(X\). We want to generalize 0-dim point and 1-dim loops to \(n\) dimension in this section.
\(I_n = [0,1]^{\otimes n} = \{(t_1,\dots, t_n)\in \mathbb R^n\ |\ t_i \in [0,1]\}\). Then \(\partial I_n = \{(t_1,\dots, t_n)\in I_n\ |\ \exists t_i =0,1\}\).
Moreover, we define a \(n\)-loop based at \(x\in X\) be a continuous map \[\gamma:I_n \to X, \mathrm{s.t.},\ \gamma|_{\partial I_n} = x.\]
Remark. This requirement is a generalization of the requirement \(\gamma(0) = \gamma(1)\) in the 1-dimensional case, i.e., loops.
Let \(\gamma, \sigma\) be two \(n\)-loops based at \(x\). Let \(\gamma(x) = \sigma(x)\) for any \(x\in \partial I_n\). Then \(\gamma\simeq \sigma\) homotopic as \(n\)-loops if there exists \(H: I_n \times I \to X\) such that \[H(t;0) = \gamma(t), H(t;1) = \sigma(t),\ \text{and}\ H(t;s) = x, \forall t\in \partial I_n, s\in I.\]
To construct group structures for \(n\)-loops, we define their composition as \[\sigma_1 *\sigma_2 = \begin{cases} \sigma_1(2t_1,t_2,\dots, t_n),\quad 0 \leq t_1\leq 1/2.\\ \sigma_2(2t_1 - 1, t_2, \dots, t_n), \frac{1}{2}\leq t_1\leq 1. \end{cases} \label{nloop}\]
Remark.
Surprisingly, \(\sigma_1 * \sigma_2 \simeq \sigma_2*\sigma_1\) for \(n \geq 2\). In other words, \((\pi_n(X,x),*)\) is an Abelian group.
Moreover, \([\sigma_1*\sigma_2]_p\) well defined, in the sense that choosing the first index in the above equations is equivalent to choosing any other coordinate in \([n]\).
Examples.
\(\pi_n(S^n) \cong \mathbb Z, n\geq 1\).
\(\pi_5(S^2) \cong \mathbb Z_2, \pi_6(S^2) \cong \mathbb Z_{12}\).
\(\pi_1(\mathbb RP^2)\cong \mathbb Z_2, \pi_n (\mathbb RP^n) \cong \mathbb Z, n\geq 2\).
(Bott periodicity of Lie groups) For \(n \geq (k+1)/2\), \[\pi_k(U(n)) \cong \pi_k(SU(n)) \cong \begin{cases} \{e\},\ k\ \text{even}\\ \mathbb Z,\ k\ \text{odd}. \end{cases}\]
For a given material, describe its magnetic feature by its magnetic moment \(\vec m(x) = \langle \sum_{i} m_i(x)\rangle\).
The free energy \[\mathcal F = \mathcal F_{grad} + \mathcal F_{potential} + \mathcal F_{ext}.\] where \[\mathcal F_{grad} = \kappa |\nabla m|^2\] and \[\mathcal F_{potential} = -\alpha(T)|m|^2 + \beta(T) |m|^4 + \frac{\alpha(T)^2}{4\beta(T)}.\] Note that we should minimize \(F = \int {\mathcal F(x)}\mathrm dx\). Suppose the gradient is zero, i.e., the magnetic moment is constant (homogeneous). If \(\alpha(T) < 0, \beta(T) > 0\), then \(|\vec m| = 0\). If \(\alpha(T) > 0, \beta (T) > 0\), then \(|\vec m| > 0\), i.e., Ferromagnetism.
We define defects as \(\vec m (x), F< \infty\) where \(\mathcal F\) is a local minimum. If \(\vec m(x)\) is regular everywhere, we call it a regular defect. If \(\vec m(x)\) has a singular point, we call it a singular defect.
For regular defects, we know when \(|x| \to \infty\), \(\mathcal F \to 0\) or otherwise the integration would not be finite. Thus, \[\lim_{|x|\to \infty} |\vec m(x)| = m_0 \propto \sqrt{\frac{\alpha}{\beta}}.\] The boundary condition for \(\vec m(x): \mathbb R^d \to \mathbb R^3\) is bounded to \(\mathbb S^2\). The solution is locally stable guaranteed by homotopy on \(\pi_{d-1}(\mathbb S^2)\).
\(\pi_0(S^2) = \pi_1(S^2) = \{0\}\), \(\pi_2(S^2) = \mathbb Z\).
\(0\)-simplex: a point \(\langle p_0\rangle\).
\(1\)-simplex: an edge: \(\langle p_0,p_1\rangle\).
\(2\)-simplex: an triangle \(\langle p_0, p_1,p_2\rangle\), but \(p_0,p_1,p_2\) do not lie on a same line.
Geometric independence: \(r+1\) points are geometrically independent, if \(p_0, \dots, p_r \subseteq r-dim\) hyperplane, but \(p_0,\dots, p_r \not \subseteq (r -1)-dim\) hyperplane.
\(\{p_0,\dots, p_r\}\) be \((r+1)\)-geometrically independent. \(p_0,\dots, p_r\in \mathbb R^m\), \(m\geq r\). A \(r\)-simplex \(\sigma^r(\langle p_0, \dots, p_r\rangle)\) is defined as \[\sigma^r = \{x\in \mathbb R^m \ |\ x = \sum_{i=1}^r c_i p_i, c_i \geq 0, \sum_{i=0}^r c_i = 1\}\]
Remark. \(\{c_i\}\) are called the barycentric coordinates of \(x\).
Let \(q\in \mathbb Z, 0\leq q \leq r\). Select \(q+1\) points from \(\{p_0,\dots, p_r\}\) and denote them by \(\{p_{i_0},\dots, p_{i_q}\}\). Then the \(q\)-simplex \(\langle p_{i_0}, \dots, p_{i_q}\rangle\) is a \(q\)-face of \(\langle p_0 ,\dots, p_r\rangle\), denoted \[\langle p_{i_0}, \dots, p_{i_q}\rangle \leq \langle p_0 ,\dots, p_r\rangle\]
\(K\) is a set of simplexes. \(K\) is a simplicial complex if
\(\sigma\in K\), \(\sigma^\prime \leq \sigma \Rightarrow \sigma^\prime \in K\)
For any \(\sigma,\sigma^\prime \in K\), either \(\sigma \cap \sigma^\prime = \varnothing\), or \(\sigma \cap \sigma^\prime \leq \sigma, \sigma \cap \sigma^\prime \leq \sigma^\prime\).
Remark. For a simplicial complex \(K\), the union of all elements in \(K\) is called a polyhedron, \(|K|:= \bigcup_{\sigma\in K} \sigma\).
Let \(X\) be a topological space. If there exists a simplicial complex \(K\), such that \(f:|K| \to X\) is a homeomorphism, then \(X\) is triangulable, and \((K,f)\) is a triangulation of \(X\).
Define orientation on \(\langle p_1,\dots, p_n\rangle\). Two simplexes have the same orientation if they differ by an even permutation.
\(K\) simplicial complex. The \(r\)-chain group on \(K\), \(C_r(K)\), is an Abelian group generated by the oriented \(r\)-simplexes of \(K\). If \(r> \dim K\), \(C_r(K) = 0\). An element in \(C_r(K)\) is called an \(r\)-chain.
Remark. \[C_r(K) = \{\sum_i z_i \sigma_i^r\ |\ z_i \in \mathbb Z, \sigma_i^r \in K\}.\]
For \(c = \sum_i c_i \sigma_i^r\), \(d = \sum_i d_i \sigma_i^r\), \(c+d = \sum_i (c_i + d_i) \sigma_i^r\).
\(C_r(K) \cong \mathbb Z^d\) for some \(d\). \(d\) is the dimension of \(C_r(K)\) and \(d=\) the number of \(r\)-simplex in \(K\).
Consider \(\dim K = 2\), \(\dim C_2(K) - \dim C_1(K) + \dim_0(K)\) is the Euler characteristic for planar graphs, which is invariant under homotopy.
Next, we define cycle groups and boundary groups that are subgroups of chain groups. We start with defining what a boundary is.
Note that the boundary of a \(r\)-simplex is a union of \((r-1)\)-simplexes it contains. We define a boundary operator \(\partial_r : C_r(K) \to C_{r-1}(K)\): \[\partial_r(\sum_i z_i \sigma_i^r) = \sum_i z_i \partial_r(\sigma_i^r)\] and for \(\sigma_i^r = (p_0,p_1,\dots, p_r)\), \[\partial_r(\sigma_i^r) = \sum_{i=0}^r (-1)^i (p_0,\dots, p_{i-1}, p_{i+1},\dots, p_r).\] By definition we know \(\sigma_r\) is a group homomorphism.
Then the kernel \(ker (\partial_r)\) is called the cycle group and the image \(Im (\partial_r)\) is called the boundary group.
An \(r\)-cycle group \(Z_r(K)\) is defined as \[Z_r(K) = Ker(\partial _r) = \{c \in C_r(K) \ |\ \partial_r c= 0\}.\]
An \(r\)-boundary group \(B_r(K)\) is defined as \[B_r(K) = Im \partial_{r+1} = \{c \in C_r(K) \ |\ c = \partial_{r+1} d, d\in C_{r+1}(K)\}.\]
\(\partial_r \circ \partial_{r+1}: C_{r+1}(K) \to C_{r-1}(K)\) is a zero map.
Therefore, \(B_r(K) \subset Z_r(K) \subset C_r(K)\).
We define a chain complex \((C_{\cdot}, d_{\cdot})\) of Abelian groups \((C_n)_{n\in \mathbb Z}\) and homomorphisms of Abelian groups \[d_n: C_{n+1} \to C_{n}\] subject to \(d_n \circ d_{n+1} = 0\) for all \(n\in \mathbb Z\).
An \(r\)-homology group of \((C_{\cdot},d)\) is defined as \[H_r(C_{\cdot}, d) = \frac{Ker(d_r: C_r \to C_{r-1})}{Im(d_{r+1}: C_{r+1} \to C_r)}\]
Let \(X\),\(Y\) be two triangulable homeomorphic topological space, and \((K,f)\), \((L,g)\) be triangulations of \(X\) and \(Y\). Then \[H_r(K) \cong H_r(L)\]
Let \(K\) be a connected simplicial complex, then \[H_0(K) \cong\mathbb Z\]
Note that \(H_r(K)\) is Abelian. The most general form of an Abelian group is \[H_r(K) \cong \mathbb Z^n \times \mathbb Z_{k_1} \times \dots \times \mathbb Z_{k_p}\] The first part is called the free part and the second is called torsion part.
If we change the coefficient in chain groups from integer to real numbers, the torsion part disappears. So \(H_r(K;\mathbb R) \cong \mathbb R^f\).
Let \(K\) be a simplicial complex. The \(r\)-th Betti number is \[b_r(K) = \dim H_r(K;\mathbb R)\] which is also the rank of the free Abelian part of \(H_r(K;\mathbb Z)\).
Let \(K\) be an \(n\)-dimensional simplicial complex, and \(I_r\) be the number of \(r\)-simplexes in \(K\). The Euler characteristics \(\chi\) is \[\chi(K) = \sum_{r=0}^n (-1)^r I_r.\]
Let \(K\) be an \(n\)-dim simplicial complex, \[\chi(K) = \sum_{r=0}^n (-1)^r b_r(K).\]
Let \(A = (a_{ij})\in M_{m\times n}(\mathbb Z)\). \(A\) is in Smith normal form if
\(a_{ij} = 0,\ \forall i\neq j\)
\(0\leq r\leq \min(m,n)\), \(a_{ii} \neq 0\) for \(i\leq r\) and \(a_{ii} = 0\) otherwise.
\(a_i = a_{ii}, 0\leq i\leq r\), then \(a_i | a_{i+1}\).
If \(A\in M_{m\times n}(\mathbb Z)\), then there exists \(U\in M_{m\times m}(\mathbb Z), V \in M_{n\times n} (\mathbb Z)\), \(\det U = \pm 1, \det V = \pm 1\), and \(D \in M_{m\times n} (\mathbb Z)\) is in Smith normal form. We have \[A = UDV.\]
\(C_{r-1}(K) \leftarrow C_r(K) \leftarrow C_{r+1}(K) \leftarrow \dots\). Let \(A\) be the matrix representation for \(\partial_{r+1}\) and \(B\) for \(\partial_r\).
Suppose \(SNF(A) = \mathrm{diag}(a_1,\dots, a_r, \mathbf 0)\), then \[H_r(K;\mathbb Z) \cong \mathrm{Ker} B/\mathrm{Im} A\cong \mathbb Z^{\dim C_r(K) - r - s}\times \mathbb Z_{a_1} \times \dots \times \mathbb Z_{a_r}\] where \(r =\mathrm{rank} A, s = \mathrm{rank} B\).