Homotopy

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Topological Invariants

In general, we are interested in classifying mathematical 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, such classification is usually difficult, and topological spaces are no exception.

The homology theories of a topological space that we defined in the previous post are topological invariants by functoriality: if two topological spaces \(X\) and \(Y\) are homeomorphic, then they are also homology equivalent. The converse does not hold in general. One might hope to find a topological invariant that completely determines a topological space, but the following remarkable result stands in the way.

Theorem 1 (Markov 1958) There is no finite algorithm that decides whether two topological manifolds of dimension at least 4 are homeomorphic.

Put naïvely, even if a topological invariant capable of distinguishing arbitrary topological spaces up to homeomorphism were to exist, there would generally be no effective method to compute it. From this perspective, topological invariants are useful not for proving that two spaces are homeomorphic, but only for showing that two spaces are not homeomorphic.

Homotopy Equivalence

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

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

holds, but neither converse holds. Moreover, homotopical equivalence is somewhat more geometrically intuitive than homology.

Definition 2 Let \(f_0,f_1:X \rightarrow Y\) be continuous maps between two topological spaces \(X,Y\). Then \(f_0\) and \(f_1\) are 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 we write \(f_0\simeq f_1\) when such a homotopy exists.

Intuitively, this means that \(f_0\) can be continuously deformed into \(f_1\). In the definition above, specifying a continuous map \(F\) is equivalent to specifying a family of continuous maps \((F(-,t))_{t\in[0,1]}\). From this viewpoint, a homotopy \(F\) between \(f_0\) and \(f_1\) 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. For any \(f\in C(X,Y)\), defining \(F(x,t)=f(x)\) yields a homotopy between \(f\) and itself.

2. Next, \(\simeq\) is symmetric. Suppose \(f_0\simeq f_1\). Then there exists a homotopy \(F\) satisfying equation (1). Define \(\tilde{F}(x,t)=F(x,1-t)\); then \(\tilde{F}\) is continuous and satisfies the two equations

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

   Hence \(f_1\simeq f_0\).

3. Finally, \(\simeq\) is transitive. Let \(f_0,f_1,f_2\in C(X,Y)\) satisfy \(f_0\simeq f_1\) and \(f_1\simeq f_2\). Then there exist homotopies \(F_0(x,t)\) and \(F_1(x,t)\) with \(F_0(x,0) = f_0(x)\), \(F_0(x,1) = f_1(x)\), \(F_1(x,0) = f_1(x)\), and \(F_1(x,1) = f_2(x)\). Define \(F(x,t)\) by

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

   Then \(F\) is a homotopy between \(f_0\) and \(f_2\).

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

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 \(g\circ f\simeq \id_{\mathbb{R}^n}\) we use the continuous map \(t\cdot\id_{\mathbb{R}^n}\) in the variable \(t\in[0,1]\) defined by

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

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 \(f_0\) and \(f_1\) are homotopic then \(C_\bullet(f_0)\) and \(C_\bullet(f_1)\) are chain homotopic. ([Homological Algebra] §Long Exact Sequence, ⁋Definition 5)

Proof

By definition, we must construct maps \(h_n:C_n(X) \rightarrow C_{n+1}(Y)\) satisfying the equation

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

The data at our disposal 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 composition

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

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

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

Then we can divide these into \((n+1)\) many \((n+1)\)-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 likewise, drawing the case \(n=2\) gives the following.

Using this decomposition we define

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

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

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

Collecting the terms with \(i=j\), in the first sum we get \(\tau\vert_{[v_0,\ldots, v_{i-1},w_i,\ldots, w_n]}\) 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 \((i-1)\)st term of the second sum cancel, 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 \(i\neq j\) amount to applying the same decomposition to the faces of \(\sigma\) rather than to \(\sigma\) itself, so their total contribution is \(-h_{n-1}(\partial_n^X\sigma)\), and thus equation (2) holds.

Therefore homotopic continuous maps induce the same map on homology. ([Homological Algebra] §Long Exact Sequence, ⁋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 homology groups \(H_\bullet(X)\) and \(H_\bullet(Y)\) are isomorphic.

Meanwhile, a constant map \(X\rightarrow Y\) to a point \(y\in Y\) 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 induced map \(H_k(X)\rightarrow H_k(Y)\) is the zero map for every \(k>0\). Thus if a continuous map \(f:X \rightarrow Y\) is homotopic to a constant map, then by Proposition 6 we know that \(H_k(f)\) is also the zero map for every \(k>0\). For this reason, a continuous map homotopic to a constant map 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, the \(k\)th homology of a contractible space is \(0\) for every \(k>0\).

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

Deformation retract

In many cases, two homotopically equivalent spaces arise from a transformation called a deformation retract, and moreover these spaces are (in contrast to the somewhat abruptly presented Definition 2) geometrically related to one another. To define this we first need the notion of retraction. ([Set Theory] §Retraction and Section, ⁋Definition 2)

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

Example 8 For instance, for the filled disc \(D^2\) in the plane 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 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 every \(n\). 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.

Yet comparing this example with the computation of the homology of \(D^2\setminus \left\{(0,0)\right\}\) in §Homology, ⁋Example 8, we see that any nontrivial homology appearing in \(D^2\setminus \left\{(0,0)\right\}\) is already present, in exactly the same form, in its subset \(S^1\). By Proposition 6 we know that homotopic continuous maps induce homotopic chain maps, and hence induce the same map on homology, so it is clear how this phenomenon should be generalized.

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

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

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

yields 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 observed that if we identify the \(1\)-simplex \(\Delta^1\) with \(I=[0,1]\), then the elements generating \(C_1(X)\) are, by definition, paths in \(X\), and in passing to homology \(H_1(X)\) we end up considering 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 identity

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

holds, we say that \(F\) is a homotopy relative to \(A\). If in Definition 9, the homotopy \(F\) is a homotopy relative to \(A\), then 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 is a homotopy relative to \(\{0,1\}\). That is, two paths \(\alpha_0,\alpha_1\) share their endpoints (so \(\alpha_0(0)=\alpha_1(0)\) 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 a slight adaptation of the proof of Proposition 3, we know that path homotopy defines an equivalence relation on the set of paths with given endpoints \(p,q\). Moreover, it is obvious that this equivalence relation is preserved under reparametrization: for any path \(\alpha:I \rightarrow X\) and any continuous map \(\varphi:I\rightarrow I\) satisfying \(\varphi(0)=0\), \(\varphi(1)=1\), 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

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

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 \(\alpha_0\ast \beta_0\sim \alpha_1\ast \beta_1\). This is immediate from the homotopy \(\alpha_t\ast \beta_t\).
2. Therefore, imposing the equivalence relation of path homotopy on \(C(I,X)\), the operation \([\alpha]\ast[\beta]=[\alpha\ast \beta]\) is well defined for each pair \(\alpha,\beta\) satisfying \(\alpha(1)=\beta(0)\).
3. Then for the constant paths \(c_{\alpha(0)}\) and \(c_{\alpha(1)}\) at the points \(\alpha(0)\) and \(\alpha(1)\) respectively, we have \([c_{\alpha(0)}]\ast [\alpha]=[\alpha]=[\alpha]\ast[c_{\alpha(1)}]\). 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 essentially the same argument, one shows that \(\ast\) is associative.

4. Moreover, inverses exist: for any path \(\alpha\), defining \(\bar{\alpha}(t)=\alpha(1-t)\), we have \([\alpha]\ast[\bar{\alpha}]=[c_{\alpha(0)}]\) and \([\bar{\alpha}]\ast[\alpha]=[c_{\alpha(1)}]\). The first equality holds because the formula

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

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

Summarizing these results, we obtain the following.

Definition 11 By the above, \(C(I,X)/{\sim}\) 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 whose objects are the points of \(X\) and whose morphisms between two points \(x,y\) are the homotopy classes of paths from \(x\) to \(y\). Let \(\Grpd\) denote the full subcategory of \(\Cat\) consisting of groupoids; then a morphism in \(\Grpd\) is simply a functor. Explicitly, given any continuous map \(f:X \rightarrow Y\), the map \(\Pi_1(f):\Pi_1(X)\rightarrow\Pi_1(Y)\) is defined on objects by \(x\mapsto f(x)\) and on morphisms by

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

That this is well defined is immediate: 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 \(f\circ\alpha_1\). Moreover, if two continuous maps \(f_0,f_1:X \rightarrow Y\) are homotopic, then 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) Namely, for any path \(\alpha:I \rightarrow X\) with starting point \(x_0\) and endpoint \(x_1\), 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 the commutativity follows from the continuous map

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

Restricting this map 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 the boundary (bottom then right, and left then top) are path homotopic. Composing with \(F\) gives a path homotopy between the two composites in the diagram.

In particular, if we consider only paths having a fixed point \(x\in X\) as both starting point and endpoint (that is, loops based at \(x\)), this amounts to considering the endomorphism monoid of \(x\) in the category \(\Pi_1(X)\); 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 therefore is equivalent to \(\Pi_1(X)\) as a category. Then, as observed above, homotopic continuous maps induce natural isomorphisms between fundamental groupoids, so the fundamental groupoid and the fundamental group are homotopy invariants.

Example 12 For instance, the fundamental groupoid \(\Pi_1(\mathbb{R}^n)\) of the space \(\mathbb{R}^n\) is the category described as follows.

• The objects of \(\Pi_1(\mathbb{R}^n)\) are exactly 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, any path \(\alpha_1:I \rightarrow \mathbb{R}^n\) from \(x_1\) to \(x_2\) is always path homotopic to the path

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

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

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

References

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