$X \simeq Y \Rightarrow \pi_0(X) \cong \pi_0(Y)$

Question

Claim: Let $X$ and $Y$ be topological spaces. Then

$$X \simeq Y \Rightarrow \pi_0(X) \cong \pi_0(Y)$$

I'm trying to prove this and I have no idea where to begin. Any hints would be helpful but I really don't want a full answer, just a jumping off point.

Thanks

Set of all connected components of a space $X$. See http://math.stackexchange.com/questions/727019/pi-0-notation — EgoKilla, Mar 26 '14 at 02:12
I think it would probably work to use the maps $f$ and $g$ which are homotopy inverse to define a map between path components of the spaces. You would have to do some work to show it is well defined. — Seth, Mar 26 '14 at 02:21
Does this answer your question? Proof of another Hatcher exercise: homotopy equivalence induces bijection — ronno, Jan 04 '20 at 20:47

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

Hint:

Prove (or use if you have already proved this) that $\pi_0$ is functorial. That is, prove that maps $f:X\longrightarrow Y$ induce maps map $f_*:\pi_0(X) \longrightarrow \pi_0(Y)$ in such a way that if we have $f:X\longrightarrow Y$ and $g:Y\longrightarrow Z$ then $(g\circ f)_* =g_*\circ f_*$, and $1_X:X\longrightarrow X$ induces $1_{\pi_0(X)}:\pi_0(X) \longrightarrow \pi_0(X)$.

Also prove that if you have $f\simeq g$ then $f_*=g_*$.

Using these facts the result follows easily.

Added later: To elaborate: Let $S^0=\lbrace -1,1\rbrace $. Let $X$ be a topological space. Fix $x_0\in X$ (the basepoint). Identify $\pi_0(X)$ as the quotient of the set of all maps $f:S^0\longrightarrow X$ sending $1$ to $x_0$, divided by the relation of based homotopy equivalence. Using this definition the functoriality of $\pi_0$ is much easier. The equivalence is not necessarily pointed, so this approach to prove the hint is not adequate, see below.

Added even later: All the maps in the original hint should be basepoint preserving for construction in the first addendum above to work, sorry (~~meaning that this "hint" doesn't work as it is~~). Meaning that the first addendum doesn't work.

(Hopefully) Final Addendum: I was going to justify that the hint does work (it really does, see the answer linked), but it turns out that this question has been asked (and answered) before: https://math.stackexchange.com/a/26188/92139

$X \simeq Y \Rightarrow \pi_0(X) \cong \pi_0(Y)$

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