Fundamental group is a homotopy invariant

Question

I am a newbie to topology and am not able to understand how to attack this problem: Any hints would be appreciated

Assuming that: $$ f \sim g \Rightarrow \pi_1(f) = \pi_1(g). $$ Prove that the fundamental group is a homotopy invariant.

Answer 1

Pick a homotopy equivalence $f: X \to Y$, with homotopy inverse $g: Y \to X$. Then by definition, $fg \sim \text{Id}_Y$ and $gf \sim \text{Id}_X$. So $\pi(fg)=\pi(f) \circ \pi(g)=\text{Id}_{\pi_1(Y)}$ and $\pi(gf)=\pi(g) \circ \pi(f)=\text{Id}_{\pi_1(X)}$, so $\pi(f)$ and $\pi(g)$ are both isomorphisms as desired.

Answer 2

This doesn't answwer the question (that was answered by Mike Miller) but I first thought that you ask about why $f\sim g$ implies $\pi_1(f)=\pi_1(g)$.

Let $F: X\times [0,1]\to Y$ be the homotopy between $f$ and $g$. Then for any loop $\varphi: S^1\to X$, the map $(s,t)\mapsto F(\varphi(s),t)$ is a homotopy between $f\circ\varphi$ and $g\circ\varphi$. The induced map $\pi_1(f)$ sends the class of $\varphi$ to the class of $f\circ\varphi$ and similarly for $g$, so $[f\circ\varphi]=[g\circ\varphi]$ for each $[\varphi]$ implies $\pi_1(f)=\pi_1(g)$. (All this can be formulated with base points as well.)