Homotopy

Topological Invariants

In general, given mathematical objects, we are interested in classifying these objects (up to isomorphism). For instance, sets are completely classified by their cardinality, and \(\mathbb{k}\)-vector spaces are completely classified by their dimension. However, in most cases such classification is not an easy task, and the same is true for topological spaces.

The homologies of a topological space that we defined in the previous post are topological invariants by functoriality. That is, if two topological spaces \(X\) and \(Y\) are homeomorphic, then they are also homology equivalent. However, in general the converse does not hold. If so, finding a topological invariant that completely determines topological spaces would be one of our hopes, but there is the following interesting result.

Theorem 1 (Markov 1958) There does not exist a finite algorithm that determines whether two topological manifolds of dimension 4 or higher are homeomorphic.

To put it naively, if there exists a topological invariant that tells whether arbitrary topological spaces are homeomorphic to each other, in general there exists no suitable method to compute this invariant. From this perspective, one could also say that topological invariants are not used to prove that two topological spaces are homeomorphic, but are useful only when showing that two topological spaces are not homeomorphic.

Homotopy Equivalence

The notion of being homotopically equivalent to be introduced in this post is likewise useful for showing that two topological spaces are not homeomorphic, and furthermore it is finer than the equivalence defined by homology. That is, the implication relation

\[X,Y\text{ homeomorphic}\implies X,Y \text{ homotopically equivalent}\implies X,Y\text{ homology equivalent}\tag{$\ast$}\]

holds, but the converse does not hold. Furthermore, being homotopically equivalent is also somewhat more geometrically intuitive compared to homology.

Definition 2 Given two topological spaces \(X,Y\), let \(f_0,f_1:X \rightarrow Y\) be continuous maps between them. Then \(f_0\) and \(f_1\) are said to be homotopic if there exists a continuous map \(F:X\times [0,1]\rightarrow Y\) such that the two equations

\[F(x,0)=f_0(x),\qquad F(x,1)=f_1(x)\tag{1}\]

hold for all \(x\in X\). In this case, we call \(F\) a homotopy between \(f_0\) and \(f_1\), and if a homotopy between \(f_0\) and \(f_1\) exists, we write this as \(f_0\simeq f_1\).

Intuitively, this means that \(f_0\) can be continuously deformed to obtain \(f_1\). In the definition above, giving a continuous map \(F\) is equivalent to giving a family of continuous maps \((F(-,t))_{t\in[0,1]}\). From this perspective, the homotopy between \(f_0\) and \(f_1\), \(F\), is also written as \((f_t)_{t\in[0,1]}\).

Proposition 3 The relation \(\simeq\) is an equivalence relation on \(C(X,Y)\).

Proof

1. First, \(\simeq\) is reflexive. This is because for any \(f\in C(X,Y)\), defining \(F(x,t)=f(x)\) defines a homotopy between \(f\) and itself.

2. And \(\simeq\) is symmetric. Assuming \(f_0\simeq f_1\), there exists a homotopy \(F\) satisfying equation (1). If we now define \(\tilde{F}(x,t)=F(x,1-t)\), then \(\tilde{F}\) is a continuous function and satisfies the two equations

   \[\tilde{F}(x,0)=f_1(x),\qquad\tilde{F}(x,1)=f_0(x)\]

   Therefore, \(f_1\simeq f_0\) holds.

3. Finally, \(\simeq\) is transitive. Suppose \(f_0,f_1,f_2\in C(X,Y)\) satisfy \(f_0\simeq f_1\) and \(f_1\simeq f_2\). Then there exist two homotopies \(F_0(x,t)\) and \(F_1(x,t)\) respectively satisfying \(F_0(x,0) = f_0(x)\) and \(F_0(x,1) = f_1(x)\), and \(F_1(x,0) = f_1(x)\) and \(F_1(x,1) = f_2(x)\). If we now define \(F(x,t)\) by the equation

   \[F(x,t) = \begin{cases} F_0(x,2t) & \text{if } 0 \leq t \leq \frac{1}{2} \\ F_1(x,2t-1) & \text{if } \frac{1}{2} \leq t \leq 1 \end{cases}\]

   then \(F\) becomes a homotopy between \(f_0\) and \(f_2\).

Basically, the equivalence relation of being homotopic is for functions as above. However, using this, we can define what it means for two topological spaces to be homotopically equivalent as follows.

Definition 4 Two topological spaces \(X,Y\) are homotopically equivalent if there exist continuous maps \(f:X\rightarrow Y\) and \(g:Y\rightarrow X\) such that \(f\circ g\simeq \id_Y\) and \(g\circ f\simeq\id_X\).

In this case, the two maps \(f,g\) satisfying the above conditions are each called a homotopy equivalence.

Example 5 For any natural number \(n\), Euclidean space \(\mathbb{R}^n\) is homotopically equivalent to the one-point space \(\{\ast\}\). A homotopy equivalence is given by

\[f:\mathbb{R}^n \rightarrow \{\ast\};\quad x\mapsto \ast,\qquad g:\{\ast\}\rightarrow \mathbb{R}^n;\quad \ast\mapsto 0\]

Then \(f\circ g=\id_{\{\ast\}}\) is obvious, and for \(g\circ f\simeq \id_{\mathbb{R}^n}\), it suffices to define the continuous map \(t\cdot\id_{\mathbb{R}^n}\) for the variable \(t\in[0,1]\) by the following formula:

\[t\cdot\id_{\mathbb{R}^n}:\mathbb{R}^n\rightarrow \mathbb{R}^n;\qquad x\mapsto tx\]

For completeness, we should establish the implication (\(\ast\)). This follows from the following more general proposition.

Proposition 6 For continuous maps \(f_0,f_1:X\rightarrow Y\), if \(f_0,f_1\) are homotopic, then \(C_\bullet(f_0), C_\bullet(f_1)\) are chain homotopic. ([Homological Algebra] §Long Exact Sequences, ⁋Definition 5)

Proof

That is, by definition, to satisfy the following equation

\[C_n(f_1)-C_n(f_0)=\partial_{n+1}^Y h_n+h_{n-1}\partial_n^X\tag{2}\]

we must construct \(h_n:C_n(X) \rightarrow C_{n+1}(Y)\), and the information we currently have is the continuous map

\[F:X\times I \rightarrow Y\]

and by definition an element of \(C_n\) is a continuous map from \(\Delta^n\) to \(X\), so the following composition

\[F\circ(\sigma\times\id_I):\Delta^n\times I \rightarrow Y\]

is well-defined. We first must use this to produce an element belonging to \(C_{n+1}(Y)\). Since the domain \(\Delta^n\times I\) of this continuous map is not an \((n+1)\)-simplex, this correspondence itself does not belong to \(C_{n+1}(Y)\). Instead, we will decompose it into a sum of \((n+1)\)-simplices and define the chain homotopy through this.

Let the vertices of the domain \(\Delta^n\times I\) on the bottom face (\(t=0\)) be \(v_0,\ldots, v_n\), and let the vertices on the top face (\(t=1\)) be \(w_0,\ldots,w_n\). The case \(n=2\) is illustrated in the figure below.

Then we can divide these into \((n+1)\) \((n+1)\)-simplices

\[[v_0,\ldots, v_n,w_n],\quad [v_0,\ldots, v_{n-1}, w_{n-1}, w_n],\quad\ldots,[v_0,w_0,\ldots, w_n]\]

and similarly, drawing the case \(n=2\) yields the following.

Using this decomposition, we can define it by the following equation:

\[h_n(\sigma)=\sum_i (-1)^iF\circ(\sigma\times\id_I)\vert_{[v_0,\ldots, v_i, w_i,\ldots, w_n]}\]

Now let us verify that this satisfies equation (2). For convenience, if we write \(\tau=F\circ(\sigma\times\id_I)\), then from the definition of the boundary map we obtain the following equation:

\[\partial_{n+1}h_n(\sigma)=\sum_{j\leq i}(-1)^i(-1)^j\tau\vert_{[v_0,\ldots,\hat{v}_j,\ldots, v_i,w_i,\ldots, w_n]}+\sum_{j\geq i}(-1)^i(-1)^{j+1}\tau\vert_{[v_0,\ldots, v_i,w_i,\ldots,\hat{w}_j,\ldots, w_n]}\]

Here, collecting only the terms with \(i=j\), in the first sum we have \(\tau\vert_{[v_0,\ldots, v_{i-1},w_i,\ldots, w_n]}\) and in the second sum \(-\tau\vert_{[v_0,\ldots, v_i,w_{i+1},\ldots, w_n]}\), so the \(i\)-th term of the first sum and the \((i-1)\)-th term of the second sum cancel each other out, leaving only the \(i=0\) term of the first sum

\[\tau\vert_{[w_0,\ldots, w_n]}=C_n(f_1)(\sigma)\]

and the \(i=n\) term of the second sum

\[-\tau\vert_{[v_0,\ldots, v_n]}=-C_n(f_0)(\sigma)\]

On the other hand, the terms with \(i\neq j\) correspond to applying the same decomposition to the faces of \(\sigma\) instead of \(\sigma\), so summing them all gives \(-h_{n-1}(\partial_n^X\sigma)\), and therefore equation (2) holds.

Therefore, homotopic continuous maps induce the same map on homology. ([Homological Algebra] §Long Exact Sequences, ⁋Proposition 6) In particular, if two spaces \(X,Y\) are homotopically equivalent and \(f:X \rightarrow Y\) and \(g:Y\rightarrow X\) are given as in Definition 4, then the homologies \(H_\bullet(X)\) and \(H_\bullet(Y)\) of the two spaces are the same.

Meanwhile, the constant map to a point \(y\in Y\), \(X\rightarrow Y\), factors through the one-point space \(\left\{y\right\}\) as \(X\rightarrow \left\{y\right\}\rightarrow Y\), so by §Homology, ⁋Proposition 11 and the functoriality of \(H_k\), the map \(H_k(X)\rightarrow H_k(Y)\) induced by the constant map is the zero map for any \(k>0\). Therefore, if some continuous map \(f:X \rightarrow Y\) is homotopic to a constant function, then by Proposition 6 we know that \(H_k(f)\) is also the zero map for any \(k>0\). For this reason, a continuous map homotopic to a constant function is called null-homotopic. In particular, if the identity map \(\id_X:X \rightarrow X\) is null-homotopic, then \(X\) is called contractible. Then by §Homology, ⁋Proposition 11 and Proposition 6 above, we know that the \(k>0\)-th homology of a contractible space is always \(0\).

In the remainder of the post, we examine homotopy equivalence and the fundamental group.

Deformation retract

In many cases, two homotopically equivalent spaces appear as the result of a deformation called a deformation retract, and moreover they also have some degree of geometric connection (in contrast to the somewhat abruptly presented Definition 2). To define this, we must first define a retraction. ([Set Theory] §Retraction and Section, ⁋Definition 2)

Definition 7 Let a topological space \(X\) and its subspace \(A\) be given. For the canonical inclusion \(\iota:A\rightarrow X\), if there exists, satisfying the equation \(r\circ\iota=\id_A\), a continuous map \(r:X\rightarrow A\), then we call \(r\) a (continuous) retraction onto the subspace \(A\), and in this case we call \(A\) a retract of \(X\).

From a set-theoretic perspective, a map \(r\) satisfying the above condition always exists, but the crucial point is that this map \(r\) is continuous.

Example 8 For instance, for the two-dimensional filled disc \(D^2\) and its boundary \(S^1\), it is known that no retraction from \(D^2\) onto \(S^1\) exists. If a retraction \(r:D^2\rightarrow S^1\) existed, then by the functoriality of \(H_n\) we would have

\[H_n(r)\circ H_n(\iota)=H_n(r\circ\iota)=H_n(\id_{S^1})=\id_{H_n(S^1)}\]

for all \(n\). In particular, \(H_n(\iota):H_n(S^1)\rightarrow H_n(D^2)\) would have to be injective. However, in §Homology, ⁋Example 8 we showed that \(H_1(D^2)\cong 0\), and following the computation for \(D^2\setminus \left\{(0,0)\right\}\) we see that \(H_1(S^1)\neq 0\), so an injective homomorphism \(H_1(\iota)\) cannot exist.

However, comparing this example with the computation in §Homology, ⁋Example 8 of the homology of \(D^2\setminus \left\{(0,0)\right\}\), we see that any nontrivial homology that can appear in \(D^2\setminus \left\{(0,0)\right\}\) consequently appears identically in its subset \(S^1\) as well. We know by Proposition 6 that homotopic continuous maps induce homotopic chain maps, and therefore that they induce the same map on homology, so it is clear how this should be generalized.

Definition 9 Let a topological space \(X\) and its subspace \(A\) be given, and let \(r:X\rightarrow A\) be a retraction. If from \(\id_X\) to \(\iota\circ r\) there exists a homotopy \(F\), then we call this a deformation retraction onto \(A\), and we call \(A\) a deformation retract of \(X\).

Then, for the retraction \(r:D^2\setminus \left\{(0,0)\right\}\rightarrow S^1\), \(x\mapsto x/\lvert x\rvert\), if we define

\[F(x,t)=t\frac{x}{\lvert x\rvert}+(1-t)x\]

this defines a homotopy from \(\id_X\) to \(\iota\circ r\). That is, \(S^1\) is a deformation retract of \(D^2\setminus \left\{(0,0)\right\}\).

Fundamental group

In §Homology, ⁋Example 8, we noted that if we view the \(1\)-simplex \(\Delta^1\) as \(I=[0,1]\), then the elements generating \(C_1(X)\) can by definition be thought of as paths in \(X\), and when obtaining the homology \(H_1(X)\) from this, we come to consider closed paths. This is essentially the same as looking at maps from \(S^1\) to \(X\). Let us treat this more rigorously.

First, for two homotopic continuous maps \(f,g:X \rightarrow Y\) and a homotopy \(F\) between them, if for a subset \(A\subseteq X\) the equation

\[F(x,t)=f(x)\qquad\text{for all $x\in A$ and $t\in[0,1]$}\]

holds, we say that \(F\) is a homotopy relative to \(A\). If in Definition 9, the homotopy \(F\) is a homotopy relative to \(A\), we call \(A\) a strong deformation retract of \(X\).

Now for any two paths \(\alpha_0,\alpha_1:I\rightarrow X\), a path homotopy between them means a homotopy relative to \(\{0,1\}\). That is, the two paths \(\alpha_0,\alpha_1\) share their endpoints (that is, \(\alpha_0(0)=\alpha_1(0)\) and \(\alpha_0(1)=\alpha_1(1)\)) and the homotopy \((\alpha_t)_{0\leq t\leq 1}\) preserves the endpoints:

\[\alpha_0(0)=\alpha_t(0)=\alpha_1(0),\qquad \alpha_0(1)=\alpha_t(1)=\alpha_1(1)\qquad\text{for all $0\leq t \leq 1$}\]

Definition 10 If there exists a path homotopy between two paths \(\alpha_0,\alpha_1:I\rightarrow X\), we say they are path homotopic and write \(\alpha_0\sim \alpha_1\).

Then, by slightly adapting the proof of Proposition 3, we know that path homotopy gives an equivalence relation on the set of paths with endpoints \(p,q\). Moreover, it is also obvious that this equivalence relation preserves reparametrization, since for any path \(\alpha:I \rightarrow X\) and, satisfying \(\varphi(0)=0\) and \(\varphi(1)=1\), any continuous function \(\varphi:I\rightarrow I\), defining

\[\alpha_t(s)=\alpha(t\varphi(s)+(1-t)s)\]

gives a path homotopy between \(\alpha_0=\alpha\) and \(\alpha_1=\alpha\circ\varphi\). Using this, we define the product of two paths by the formula

\[(\alpha\ast \beta)(s)=\begin{cases}\alpha(2s)&0\leq s \leq 1/2\\ \beta(2s-1)&1/2\leq s \leq 1\end{cases}\tag{$\ast\ast$}\]

For this to be a continuous path, \(\alpha\) and \(\beta\) must of course satisfy \(\alpha(1)=\beta(0)\). Then the following properties hold.

1. If \(\alpha_0\sim \alpha_1\) and \(\beta_0\sim \beta_1\), and \(\alpha_0\ast \beta_0\) is well-defined, then \(\alpha_1\ast \beta_1\) is also well-defined and \(\alpha_0\ast \beta_0\sim \alpha_1\ast \beta_1\) holds. This is obvious by considering the homotopy \(\alpha_t\ast \beta_t\).
2. Therefore, if we give an equivalence relation on \(C(I,X)\) by path homotopy, through the formula \([\alpha]\ast[\beta]=[\alpha\ast \beta]\), whenever \(\alpha(1)=\beta(0)\) is satisfied, for each \(\alpha,\beta\), the operation of \([\alpha]\) and \([\beta]\) is well-defined.
3. Then, for the constant paths staying at the points \(\alpha(0)\) and \(\alpha(1)\), \(c_{\alpha(0)}\) and \(c_{\alpha(1)}\), \([c_{\alpha(0)}]\ast [\alpha]=[\alpha]=[\alpha]\ast[c_{\alpha(1)}]\) holds. Roughly speaking, this is because path homotopy preserves reparametrization, so in formula (\(\ast\ast\)) we may replace \(1/2\) by any number between \(0\) and \(1\), and letting this number approach \(0\) or \(1\) gives the desired homotopy. By almost the same logic, one can show that \(\ast\) is associative.

4. Moreover, inverses also exist: for any path \(\alpha\), if we define \(\bar{\alpha}(t)=\alpha(1-t)\), then \([\alpha]\ast[\bar{\alpha}]=[c_{\alpha(0)}]\) and \([\bar{\alpha}]\ast[\alpha]=[c_{\alpha(1)}]\) hold. The first equality holds because the following formula, which traces \(\alpha\) only up to the parameter \(1-t\) and then returns,

   \[F(s,t)=\begin{cases}\alpha(2s(1-t))&0\leq s \leq 1/2\\ \alpha(2(1-s)(1-t))&1/2\leq s \leq 1\end{cases}\]

   gives a path homotopy between \(\alpha\ast \bar{\alpha}\) and \(c_{\alpha(0)}\), and the second equality is obtained by applying this to \(\bar{\alpha}\).

Summarizing these results gives the following.

Definition 11 By the above results, \(C(I,X)/{\sim}\) forms a groupoid; we call this the fundamental groupoid of \(X\) and denote it by \(\Pi_1(X)\). ([Category Theory] §Category, ⁋Definition 11)

That is, for any space \(X\), the fundamental groupoid \(\Pi_1(X)\) is the category having the points of \(X\) as objects and the homotopy types of paths from \(x\) to \(y\) as morphisms for two points \(x,y\). If we let the full subcategory of \(\Cat\) consisting of groupoids be \(\Grpd\), a morphism in \(\Grpd\) is simply a functor. Explicitly, given any continuous map \(f:X \rightarrow Y\), \(\Pi_1(f):\Pi_1(X)\rightarrow\Pi_1(Y)\) is defined on objects by \(x\mapsto f(x)\), and on morphisms by the composition

\[\Pi_1(f)([\alpha])=[f\circ\alpha]\]

and that this is well-defined is trivial, since for two paths \(\alpha_0\sim\alpha_1\), if \(\alpha_t\) is a path homotopy between them, then \(f\circ \alpha_t\) is a path homotopy between \(f\circ\alpha_0\) and \(f\circ\alpha_1\). Furthermore, if two continuous maps \(f_0,f_1:X \rightarrow Y\) are homotopic, there exists a natural isomorphism between the two functors \(\Pi_1(f_0)\) and \(\Pi_1(f_1)\) induced by them. ([Category Theory] §Natural Transformations, ⁋Definition 1) That is, with \(x_0,x_1\) as starting point and endpoint respectively, for any path \(\alpha:I \rightarrow X\), the following diagram

commutes (up to path homotopy). Here, \(f_t(x_0)\) and \(f_t(x_1)\) are the paths arising from the homotopy \((f_t)_{0\leq t\leq 1}\), and commutativity follows from the continuous function

\[F(s,t)=f_t(\alpha(s))\]

Restricting this function to the four sides of the domain \(I\times I\) yields \(f_0\circ\alpha\), \(f_1\circ\alpha\), \((f_t(x_0))_{0\leq t\leq 1}\), and \((f_t(x_1))_{0\leq t\leq 1}\) respectively, and since \(I\times I\) is convex, the two paths from \((0,0)\) to \((1,1)\) along its boundary (the bottom side then the right side, and the left side then the top side) are path homotopic to each other. Composing this with \(F\) gives a path homotopy between the two compositions in the diagram.

In particular, if we consider only paths having a fixed point \(x\in X\) as starting point and endpoint (that is, loops with base point \(x\)), this is equivalent to considering the endomorphism monoid in the category \(\Pi_1(X)\) at \(x\), and since \(\Pi_1(X)\) is a groupoid, this is in fact an automorphism group. We denote this by

\[\pi_1(X,x)=\Aut_{\Pi_1(X)}(x)\]

and if \(X\) is path-connected, this group does not depend on the choice of \(x\). We call this the fundamental group of \(X\). In this case, \(\pi_1(X,x)\) is a skeleton of \(\Pi_1(X)\) ([Category Theory] §Natural Transformations, ⁋Definition 4) and is therefore equivalent to \(\Pi_1(X)\) as a category. Then, as seen above, homotopic continuous maps induce natural isomorphisms between fundamental groupoids, so we see that the fundamental groupoid and the fundamental group are homotopy invariants.

Example 12 For example, for the space \(\mathbb{R}^n\), the fundamental groupoid \(\Pi_1(\mathbb{R}^n)\) is the category consisting of the following data:

• The objects of \(\Pi_1(\mathbb{R}^n)\) are precisely the points of \(\mathbb{R}^n\).

• For any \(x_1,x_2\in \mathbb{R}^n\), there exists a unique morphism from \(x_1\) to \(x_2\). That is, starting from \(x_1\) and going to \(x_2\), any path \(\alpha_1:I \rightarrow \mathbb{R}^n\) is always path homotopic to the following path

  \[\alpha_0:t\mapsto (1-t)x_1+tx_2\]

  This can be easily verified by setting \(\alpha_t=(1-t)\alpha_0+t\alpha_1\).

Therefore, for any \(x\), \(\pi_1(\mathbb{R}^n,x)\) is the trivial group.

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References

[Hat] A. Hatcher, Algebraic Topology. Cambridge University Press, 2022.
[Mun] James Munkres, Topology. Prentice Hall, 2000.

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