Let $X$ and $Y$ be topological spaces. If $f\colon X\to Y$ is a homeomorphism, then it induces an isomorphism $f_\sharp\colon\pi_1(X,x_0)\to \pi_1(Y,f(x_0))$. All good.
As far as I know, the result still holds if we weaken "homeomorphism" to "homotopy equivalence", but I am having trouble checking that. I know that $(f_2\circ f_1)_\sharp = (f_2)_\sharp\circ (f_1)_\sharp$, $({\rm id}_X)_\sharp = {\rm id}_{\pi_1(X,x_0)}$ and that $f_1\simeq f_2 ~ {\rm rel}\,\{x_0\}$ implies $(f_1)_\sharp = (f_2)_\sharp$.
If $f\colon X \to Y$ is a homotopy equivalence, then we have $g\colon Y\to X$ with $g\circ f \simeq {\rm id}_X$ and $f\circ g\simeq {\rm id}_Y$.
My problems are:
- I can't guarantee that $g(f(x_0))=x_0$;
- These homotopies don't need to be ${\rm rel}\,\{x_0\}$ and ${\rm rel}\,\{f(x_0)\}$, respectively.
If I have these two assumptions, I can make: $${\rm id}_{\pi_1(X,x_0)} = ({\rm id}_X)_\sharp = (g\circ f)_\sharp = g_\sharp\circ f_\sharp$$and repeat this for $f_\sharp\circ g_\sharp$.
How can I go around these problems?
Let $H$ be a homotopy from $g\circ f$ to ${\rm id}_X$ and take a path $\gamma(t) = H(x_0, t)$, i.e. $\gamma(0) = g(f(x_o))$ and $\gamma(1) = x_0$. Consider
$\pi_1(X, x_0) \to \pi_1(Y, f(x_0)) \to \pi_1(X, g \circ f(x_0)) \to π_1(X, x_0)$,
where the first map is $f_{\sharp}$, the second $g_{\sharp}$ and the third is the one induced by $\gamma$. Hence $[\lambda] \in \pi_1(X, x_0)$ gets mapped to $[\gamma^{-1} g f \lambda \gamma]$. One immediately writes down (with the help of $H$) a homotopy from $\lambda$ to $g f \lambda$. Use it to write down a homotopy from $\lambda$ to $\gamma^{-1} g f \lambda \gamma$ concluding that the composition of those maps is indeed the identity.