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
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