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I'm doing stuck with the first half of exercise 12 on page 19 in Hatcher:


Show that a homotopy equivalence $f : X \rightarrow Y$ induces a bijection between the set of path-components of $X$ and the set of path-components of $Y$. and that $f$ restricts to a homotopy equivalence from each path-component of $X$ to the corresponding path-component of $Y$

By the definition of homotopy equivalence, there exists $g$ such that $gf \simeq id_X$ and $fg \simeq id_Y$. So $g$ is a two-sided inverse of $f$, therefore $f$ is a bijection.

I'm stuck now because I don't know what there is to prove to show that "$f$ restricts to a homotopy equivalence from each path-component of $X$ to the corresponding path-component of $Y$".

Is there anything to prove? If $f$ is a homotopy equivalence, is it not obvious that any restriction is also a homotopy equivalence? Many thanks for helping me, I really appreciate it.


So here is an attempt of proof that $f^\ast:\pi_0(X) \rightarrow \pi_0(Y)$ is a bijection:

Define $f^\ast$ as $f^\ast (A) := B$ where $B$ is the path-(connected)-component that contains $f(A)$ and $A \in \pi_0 (X)$.

Claim: $f^\ast$ is bijective


(i) injective:

Let $f(A) = f(B)$ where $A, B \in \pi_0(X)$.

Then $f(A) \cup f(B)$ is a path-(connected)-component.

Then $f(A \cup B)$ is also a path-component.

By the definition of homotopy equivalence, there exists $g$ (continuous) such that $gf \simeq id_X$ and $fg \simeq id_Y$. So $g(f(A \cup B)) = A \cup B$ is path-connected.

$\implies A = B$.

In the proof above I used the following claim:

Claim: $A$ path-connected, $f$ continuous $\implies f(A)$ path-connected.

Proof: Assume $f(A)$ not path-connected, then $f(A)$ not connected, then $g(f(A))=A$ not connected $\implies$ contradiction.

(ii) surjective:

Let $B \in \pi_0(Y)$. Then for $A := g(B)$, $f(A) = B$.

Can you tell me if my proof is correct? I really appreciate your help!

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You're using $g(f(A)) = A$ twice. Why is that? Also, the implication $f(A)$ not path-connected so $f(A)$ not connected is false (connectedness doesn't imply path-connectedness). To see that the image of a path-connected space under a continuous function is path-connected, pick two points $x,y \in f(A)$ There are points $a,b \in A$ such that $f(a) = x$ and $f(b) = y$. Since $A$ is path-connected, there is a path $\gamma:[0,1] \to A$ connecting $a$ and $b$. But then $f \circ \gamma$ will be a path connecting $x = f(a)$ and $y = f(b)$. – t.b. Mar 22 '11 at 9:44
@Theo: many thanks, you're truly helpful. I just noticed your edit of your answer. I haven't read it but it looks as if you solved the exercise for me. I guess I was being too slow... : / I'll do the exercise again after reading your answer. Many thanks again! – Rudy the Reindeer Mar 22 '11 at 11:03
@Theo: that I hadn't accepted your (first) answer didn't mean I wasn't going to! I left this question "unaccepted" because I personally wasn't done with this exercise, not because I wasn't satisfied with your answer. – Rudy the Reindeer Mar 22 '11 at 11:07
Don't worry you asked the question, so it's up to you to decide if an answer is acceptable or not. The whole thing turned out to be a bit long-winded, but I tried to formulate it in such a way that you see what you have to do in other situations (for instance the corresponding fact about $\pi_{n}(X)$):1. Prove that a continuous map $f:X \to Y$ yields a homomorphism $f_{\ast}: \pi_{n}(X) \to \pi_n(Y)$. 2. Prove that $(\operatorname{id}_{X})_\ast: \pi_{n}(X) \to \pi_{n}(Y)$ and $f_{\ast}g_{\ast} = (fg)_\ast$. 3. Prove $f \simeq f'$ implies $f_{\ast} = f'_\ast$. The rest is formal manipulation. – t.b. Mar 22 '11 at 11:19
I meant $(\operatorname{id}_X)_\ast = \operatorname{id}_{\pi_{n}(X)}$ but I ran out of characters. – t.b. Mar 22 '11 at 11:21
up vote 10 down vote accepted

Be careful: A homotopy equivalence is by no means a bijection in general: Let $X = \mathbb{R}$ and $Y = \{\ast\}$. Also, a homotopy equivalence need not restrict to a homotopy equivalence of subspaces. For instance, the orthogonal projection of $\mathbb{R}^2$ onto the $x$ axis is a homotopy equivalence. However, it maps the unit circle to the interval $[-1,1]$ and these two spaces certainly aren't homotopy equivalent (but this isn't completely obvious). Even simpler, take the subspace $([-1,1] \times \{0\}) \cup ([-1,1] \times \{1\})$ of $\mathbb{R}^2$ and its image $[-1,1]$ under the projection: you'll get a contradiction to your exercise.

The expressions $gf \simeq \operatorname{id}_{X}$ and $fg \simeq \operatorname{id}_{Y}$ mean that $gf$ and $fg$ are homotopic to the respective identities. (btw: it is more customary to use $\simeq$ \simeq instead of $\cong$ for "homotopic")

A continuous map sends path components into path components (because it sends paths to paths). The path component of a point $x$ is sent into the path component of its image $f(x)$ and the path component of $f(x)$ is sent into the path component of $gf(x)$. But as $gf \simeq \operatorname{id}_{X}$, a homotopy $H: X \times [0,1] \to X$ such that $H(\cdot,0) = gf$ and $H(\cdot,1) = \operatorname{id}_{X}$ gives us the path $\gamma(t) = H(x,t)$ connecting $gf(x)$ with $x$, so the path component of $gf(x)$ is the same as the path component of $x$. I let you finish up the argument yourself.

Added in response to the edited question.

Let $[x]$ be the path component of $x$. Let me write $f_{\ast}: \pi_{0}(X) \to \pi_{0}(Y)$ instead of $f^\ast$. By definition $f_{\ast}[x] = [f(x)]$ (check that this is well-defined!). In my last paragraph above I argued that $[gf(x)] = [x]$ so $g_\ast f_\ast [x] = [x]$ (check that $(gf)_\ast = g_\ast f_\ast$!). In other words $g_\ast f_\ast = (\operatorname{id}_{X})_{\ast} = \operatorname{id}_{\pi_{0}(X)}$ (check that $(\operatorname{id}_X)_\ast = \operatorname{id}_{\pi_{0}(X)}$!). In particular, $f_{\ast}$ is injective and $g_\ast$ is surjective. By symmetry we have $f_\ast g_\ast = \operatorname{id}_{\pi_{0}(Y)}$, so $f_{\ast}$ and $g_\ast$ are mutually inverse bijections.

A bit later

Maybe it's better to start from scratch.

Define an equivalence relation on $X$ by $x \sim x'$ if and only if there is a path connecting $x$ and $x'$. The equivalence classes of $\sim$ are precisely the path components of $X$. In other words, $\pi_{0}(X) = X/\!\!\sim$. Write $[x]$ for the $\sim$-equivalence class of $x$ (so $[x]$ is the path component of $x$).

Fact 1: A continuous map $f: X \to Y$ sends path components into path components. In other words, if $x \sim x$ then $f(x) \sim f(x')$.

Indeed, if $\gamma: [0,1] \to X$ is a path with $\gamma(0) = x$ and $\gamma(1) = x'$ then $f \circ \gamma: [0,1] \to Y$ is a path and $f(\gamma(0)) = f(x)$ and $f(\gamma(1)) = f(x')$, so $f(x) \sim f(x')$.

Again in other words Fact 1 tells us that $f$ yields a map $f_{\ast}: \pi_{0}(X) \to \pi_{0}(Y)$. Explicitly, \[ f_{\ast}([x]) = [f(x)]. \] This is well-defined because for $x \sim x'$ we have $f(x) \sim f(x')$, so $[f(x)] = [f(x')]$.

From this description we see that $(\operatorname{id}_{X})_{\ast}([x]) = [\operatorname{id}_{X}(x)] = [x] = \operatorname{id}_{\pi_{0}(X)}([x])$, so $(\operatorname{id}_{X})_{\ast} = \operatorname{id}_{\pi_{0}(X)}$. Also $(gf)_{\ast}([x]) = [gf(x)] = g_{\ast}([f(x)]) = g_{\ast}f_{\ast}([x])$, so $(gf)_{\ast} = g_{\ast}f_{\ast}$.

Since the two identities I've just proven are so important, let me state them again for emphasis:

Fact 2: For the identity $\operatorname{id}_{X}: X \to X$ and any two maps $f:X \to Y$ and $g: Y \to Z$ we have $$\displaystyle (\operatorname{id}_{X})_{\ast} = \operatorname{id}_{\pi_{0}(X)}: \pi_{0}(X) \to \pi_{0}(X) \qquad\text{and}\qquad (gf)_{\ast} = g_\ast f_\ast : \pi_{0}(X) \to \pi_{0}(Z) $$

Fact 3: If $f, f': X \to Y$ are homotopic, $f \simeq f'$, then $f_{\ast} = f'_{\ast}: \pi_{0}(X) \to \pi_{0}(Y)$.

Indeed, pick a homotopy $H: X \times [0,1] \to Y$ such that $f = H(\cdot,0)$ and $f' = H(\cdot,1)$. Since $t \mapsto H(x,t)$ is a path connecting $f(x)$ to $f'(x)$ we see that $f(x) \sim f'(x)$, so $f_{\ast}[x] = [f(x)] = [f'(x)] = f'_\ast[x]$.

Now we are finally in shape to solve the exercise. Let $f: X \to Y$ and $g: Y \to X$ be such that $gf \simeq \operatorname{id}_{X}$ and $fg \simeq \operatorname{id}_{Y}$. Then combining facts 2 and 3 we have that $$\displaystyle (\operatorname{id}_{X})_\ast = g_{\ast} f_{\ast} \qquad \text{and}\qquad (\operatorname{id}_{Y})_\ast = f_{\ast} g_{\ast} $$ and as $(\operatorname{id}_X)_\ast = \operatorname{id}_{\pi_{0}(X)}$ we see that $f_{\ast}$ and $g_{\ast}$ are mutually inverse bijections.

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It is interesting to see how much easier something can be using good notation (yours instead of mine). – Rudy the Reindeer Mar 23 '11 at 16:34
@Matt: Thanks, I guess that is a major lesson to be learned. It's definitely a matter of experience and taste to choose good and self-evident notation. Many things just prove themselves that way. And of course the notation is not mine, it's completely standard. – t.b. Mar 24 '11 at 10:32

If $D$ is a path component of $X$, then $f$ restricts to a map $D \to Y$ which lands in some path component $E$ of $Y$. If you want to claim that $f$ automatically restricts to a homotopy equivalence from $D$ to $E$, you have to show first that $g(E) \subseteq D$ and second that the induced homotopy $H$ between $gf$ and $\text{id}_D$ never has image outside of $D$ (and the corresponding statement for $E$). This is stronger than what you're being asked to prove; all you need to show is that $f$ and $g$ induce inverse maps on path components.

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but isn't $g = f^{-1}$ and therefore $g(E)=f^{-1}(E)= f^{-1}(f(D)) = D$? – Rudy the Reindeer Mar 10 '11 at 11:08
@Matt: you don't seem to have understood the definition of homotopy equivalence. $g$ is not an inverse of $f$; it is only an inverse up to homotopy. In particular, you claim that $f$ is a bijection; this is false in general. – Qiaochu Yuan Mar 10 '11 at 11:11
You are right : ( I have to let things sink in I think, it's all too new. – Rudy the Reindeer Mar 10 '11 at 12:11

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