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?



Are there homotopy equivalences that are not weak homotopy equivalences?
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.


