# Homotopy inverses need not induce inverse homomorphisms

Let $f:X \rightarrow Y$ and $g : Y \rightarrow X$ be homotopy inverses, ie. $f \circ g$ and $g\circ f$ are homotopic to the identities on $X$ and $Y$. We know that $f_*$ and $g_*$ are isomorphisms on the fundamental groups of $X$ and $Y$. However, it is my understanding that they need not be inverse isomorphisms. Is there an explicit example where they are not?

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This can only happen if $f$, $g$ and the homotopies $f\circ g \simeq id$ and $g\circ f \simeq id$ do not preserve base points. –  Alexander Thumm Mar 29 '12 at 15:07

If $f,g$ are pointed maps (which is necessary so that $f_*,g_*$ make sense): No, they induce inverse homomorphsism.
Homotopic maps induce the same maps on homotopy groups, in particular fundamental groups. This means that we have a functor $\pi_1 : \mathrm{hTop}_* \to \mathrm{Grp}$. Every functor maps two inverse isomorphisms to the corresponding two inverse isomorphisms.
You can make sense of $f_*: \pi_1(X,x_0) \to \pi_1(Y,y_0)$ and $g_*: \pi_1(Y,y_0) \to \pi_1(X,x_1)$ with $y_0 = f(x_0)$ and $x_1 = g(y_0)$. Now you can identify $\pi_1(X,x_0)$ and $\pi_1(X,x_1)$ by choosing a path $\gamma$ from $x_0$ to $x_1$. Take for example $y(t) = H(x_0,t)$, where $H: id \simeq g \circ f$. Now the same thing can be done in the other direction, so you get some "induced" maps. The problem with this is, that these maps depend on paths chosen and are by no means natural/functorial. –  Alexander Thumm Mar 31 '12 at 12:07
In my experience most introductions to the concept of fundamental groups do not emphasize enough the importance of base points. One often encounters informal statements like "$\pi_1(S^1) = \mathbb Z$" which are not properly explained. –  Alexander Thumm Mar 31 '12 at 12:16