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This is question 7 on page 366 from section 58 of Munkres Topology:

Let $A$ be a subspace of $X$, let $j: A \to X$ the inclusion map, $f:X \to A$ continuous. Suppose there is a homotopy $H$ between $j \circ f$ and the identity on $X$.

a) Show if $f$ is a retraction, then $j_{*}$ is an isomorphism.

b) Show if $H$ maps $A \times I$ into $A$, then $j_{*}$ is an isomorphism.

c) Give an example where $j_{*}$ is not an isomorphism.

For part a, the book essentially shows how to do it for deformation retractions. Is the idea the same for general retractions? (compose, note one direction is the identity, the other direction is homotopic to the identity, etc...)

I'm not sure how to do part b. If $A$ is pointwise fixed, this is just equivalent to the notion of a deformation retraction, but I am not sure how to do it if the set itself is the only thing fixed.

For part c, I can think of some weird examples, but are there any very natural ones?

Johnny Apple
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2 Answers2

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In both a and b, we have $j_*$ is surjective because for any element in $\pi_1(X)$ and loop in $X$ representing that element, the homotopy $H$ gives an explicit loop contained in $A$ which represents the same element in $\pi_1(X)$.

For injective, say we have two loops $\gamma_1$ and $\gamma_2$ contained in $A$ which represent the same element in $\pi_1(X)$. Let $h:S^1\times I \to X$ be a homotopy between them. We now split in cases depending on the condition, trying to prove that $\gamma_1$ and $\gamma_2$ are homotopic in $A$:

  • In a, we have that $f$ is a retraction, that is, constant on $A$. That means that applying $f$ to the loops $\gamma_1$ and $\gamma_2$ doesn't change them. Now $f\circ h$ is a homotopy between the circles completely contained in $A$, so $\gamma_1$ and $\gamma_2$ represent the same element in $\pi_1(A)$.

  • In b, the homotopy $H$ keeps $A$ inside $A$, but $f$ might not keep the loops fixed. In this case, some stitching is required. The map $f \circ h$ is a homotopy in $A$ between $f(\gamma_1)$ and $f(\gamma_2)$. Now by $H$, the loops $\gamma_1$ and $f(\gamma_1)$ are homotopic inside $A$, and the same for $\gamma_2$ and $f(\gamma_2)$. So by stitching together three different homotopies between four loops we have found that $\gamma_1$ and $\gamma_2$ are homotopic in $A$.

As for a counterexample, we have a few hints, the most important one is that no matter what we try, $j_*$ will end up being surjective. Now, the easiest surjective, non-injective group homomorphism is the trivial homomorphism $\Bbb Z \to 0$. So, in that spirit, let $X = \Bbb R^2$, $A$ an annulus and $f$ the constant map to some point in $A$.

Arthur
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  • Ok...I had a similar idea on part b. Thanks alot! – Johnny Apple Jun 26 '14 at 20:54
  • @Arthur.. I don't understand the idea here. For (b) I have $j_{}: \pi_{1}{A}\to \pi_{1}(X)$. All I need to show is that $j_{}$ is one to one. So I consider $j_{}(\gamma_1)=j_{}(\gamma_2)$. This means that $[j\circ \gamma_1]=[j\circ \gamma_2]$. If I can a find a homotopy between $f\circ \gamma_1$ and $f\circ \gamma_2$, then I will be done. – tattwamasi amrutam Feb 22 '16 at 07:08
  • @tattwamasiamrutam You need a homotopy between $\gamma_1$ and $\gamma_2$ that is contained in $A$. What you have is a homotopy in $A$ between $f\circ \gamma_1$ (which is not $\gamma_1$) and $f\circ \gamma_2$ (same), and you have a homotopy between $\gamma_1$ and $\gamma_2$ that is contained in $X$, not $A$. I solve this by using the former (homotopy between $f\circ \gamma_1$ and $f\circ\gamma_2$) together with a homotopy from $\gamma_1$ to $f\circ \gamma_1$ and from $f\circ\gamma_2$ and $\gamma_2$ (both given by $H$, since it specifically keeps $A$ inside $A$). – Arthur Feb 22 '16 at 10:23
  • Why is there a homotopy between $\gamma_1$ and $\gamma_2$?? – tattwamasi amrutam Feb 22 '16 at 11:20
  • @tattwamasiamrutam Because you have assumed that $[j\circ \gamma_1]=[j\circ \gamma_2]$, and that is exactly what that means: There is a homotopy in $X$ between the two loops. Applying $f$ gives you a homotopy between $f\circ \gamma_1$ and $f\circ\gamma_2$ that is contained in $A$, but as I said, this is not the same as a homotopy in $A$ from $\gamma_1$ to $\gamma_2$. – Arthur Feb 22 '16 at 11:24
  • Why is $f:S^1 \times I \to X$?? – tattwamasi amrutam Feb 22 '16 at 11:28
  • @tattwamasiamrutam I assume you mean $h$. The answer is because we've assumed that $[j\circ \gamma_1]=[j\circ \gamma_2]$, and therefore that there is a homotopy in $X$ that connects the two loops. That homotopy is called $h$, and we have no guarantees that said homotopy will keep everything neatly inside $A$. Therefore we need to use $h$ and $f$ and $H$ all together to construct a homotopy that stays inside $A$. Only then can we conclude that $[\gamma_1] = [\gamma_2]$, and thus that $j_*$ is injective. – Arthur Feb 22 '16 at 11:47
  • @Arthur Yeah I meant $h$. I understood the idea here. Since $j\circ \gamma_1(t)=\gamma_1(t)$ and similarly $j\circ \gamma_2(t)=\gamma_2(t)$, there exists $h: I\times I \to X$ , where $h$ is the homotopy between them. This gives that $f \circ h$ is the homotopy between $f\circ \gamma_1$ and $f\circ \gamma_2$ in $A$. But $H$ is the homotopy between $\gamma_i$ and f \circ \gamma_i$ for $i=1,2$. Thus $\gamma_i$ are homotopic to each other. – tattwamasi amrutam Feb 22 '16 at 12:03
  • @tattwamasiamrutam Except for a missed $-sign in your comment, yes, that's basically it. All that is needed to prove that two loops that we know are homotopic in $X$ are, in fact, homotopic in $A$. – Arthur Feb 22 '16 at 12:13
  • @Arthur I'm confused by one thing in your proposed solution. How do we compare $\gamma_1$ and $f \gamma_1$ in the fundamental group; if $\gamma_1$ is a loop based at say $a_0$, then $f \gamma_1$ is a loop based at $f(a_0)$; these may not be the same point? Without knowing X is path connected or having an explicit path from $a_0$ to $f_(a_0)$, how can we speak of homotopy between loops based at possibly different points? – Omnivium Feb 22 '16 at 18:18
  • @Omnivium s is stated in the question, $H$ is a homotopy between $j\circ f$ and the identity on $X$, and therefore gives you a path from $a_0$ to $f(a_0)$. – Arthur Feb 22 '16 at 18:22
  • @Arthur: for surjectivity: I understand that you aim to show that for a loop $\gamma$ in $X$ we have $[\gamma] = [f \circ \gamma]$. This works for (a), but not for (b), since not necessarily $f(a) = a$ in (b). So they are loops from a different starting point right? – Sigurd May 16 '18 at 15:00
  • @Sigurd Yes, that is true. I don't remember exactly what formulations Munkres uses, but one resolution is that $t\mapsto H(a,t)$ gives a path from $a$ to $f(a)$. Conjugating by this path is an isomorphism from $\pi_1$ at $a$ to $\pi_1$ at $f(a)$. – Arthur May 16 '18 at 16:03
  • Cool thanks. Indeed I had something like that in mind. I think the notation used in Munkres for this is $\widehat{\alpha}$. – Sigurd May 16 '18 at 21:11
3

For the part (a), there is another quicker way to solve it.

Recall that

Let $f:(X, x_{0})\longrightarrow (Y, y_{0})$ be continuous. Show that if $f$ is a homotopy equivalence, then $$f_{*}:\pi_{1}(X, x_{0})\longrightarrow\pi_{1}(Y,y_{0}),$$ is an isomorphism.

Then all you need to do is to show $j$ is a homotopy equivalence.

Proof:

Let $a$ be the base point. Then since $j$ is an inclusion, $j(a)=a$. Therefore, $$j:(A,a)\longrightarrow (X,a)$$ and hence the induced homomorphism is $$j_{*}:\pi_{1}(A,a)\longrightarrow\pi_{1}(X,a).$$

Now, since $f$ is a retraction of $X$ onto $A$ and $j$ is an inclusion, we must have $f\circ j=Id_{A}$. But by hypothesis we also have $j\circ f\sim Id_{X}$.

Thus, $j:(A,a)\longrightarrow (X,a)$ is a homotopy equivalence, and thus $j_{*}$ is an isomorphism.

For (c)

The basic idea is the same but using $\mathbb{S}^{1}$ seems more straightforward, and I add more details to help you understand.

let $X=\mathbb{R}^{2}$, $A=\mathbb{S}^{1}$ and $f$ the constant map that maps everything in $X$ to a point in $c\in A$.

Then since $\mathbb{R}^{2}$ is convex, we can define a homotopy via the affine combination, i.e. the map $H:X\times [0,1]\longrightarrow X$ defined by $H(x,t):=tx+(1-t)c$, which is clearly continuous.

Then $H(x,0)=c=j(c)=j\circ f(x)$ since $j$ is just an inclusion map, $H(x,1)=x=Id_{X}(x)$.

Thus, $H$ is the desired homotopy such that $j\circ f\sim Id_{X}$.

Now, let the base point be $1$, then $j:(\mathbb{S}^{1},1)\longrightarrow (\mathbb{R}^{2}, 1)$ and thus the induced homomorphism is $$j_{*}:\pi_{1}(\mathbb{S}^{1},1)\longrightarrow\pi_{1}(\mathbb{R}^{2},1).$$

Recall that $\pi_{1}(\mathbb{S}^{1}, 1)\cong\mathbb{Z}$, and since $\mathbb{R}^{2}$ is convex, then every loop in the $\pi_{1}(\mathbb{R}^{2},1)$ is path-homotopic to the constant loop $\epsilon_{1}$ via a path-homotopy defined via affine combination and hence $\pi_{1}(\mathbb{R}^{n},1)$ is trivial.

Thus, $j_{*}$ is actually a homomorphism from $\mathbb{Z}$ to $e_{\pi_{1}(\mathbb{R},1)}$, as desired.

JacobsonRadical
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