I can imagine a map $f: X\to Y$ which is a homotopy equivalence of unpointed spaces, but which is not a homotopy equivalence of pointed spaces, no matter what basepoint is chosen. That being the case, I don't see why $f$ would have to be a weak homotopy equivalence.

More detail: by choosing $x\in X$, and its image $y\in Y$ as basepoints, we get a pointed map and an induced map on homotopy groups. To be a weak homotopy equivalence, this needs to be an isomorphism (and one point is as good as any point if $X$ is path connected). But this (hypothetical) pointed map is not invertible in the homotopy category of pointed spaces, so why should it induce an isomorphism?


1 Answer 1


f is always a weak homotopy equivalence. This is related to the following assertion: Let $g\colon X\to X$ be homotopic to the identity. Then for any $x_0\in X$. $g$ induces an isomorphism $g_*\colon\pi_*(X,x_0)\to \pi_*(X, g(x_0))$. This is so because $g_*$ is the same as conjugation by the path $H(x_0\times I)$, where $H:X\times I \to X$ is the homotopy of $g$ with the identity. Conjugation by a path induces an isomorphism.

Now let $h$ be a homotopy inverse of $f$ and consider the homomorphisms

$$\pi_*(X, x_0)\to \pi_*(Y, f(x_0))\to \pi_*(X, hf(x_0)) \to \pi_*(Y, fhf(x_0))$$

the first and the third being induced by $f$ and the second by $h$. The composition of the first two is an isomorphism and so is the composition of the last two. It follows that all three homomorphisms are isomorphisms.


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