I've been trying to show the following:
EDIT: Let $X$ and $Y$ be compactly generated topological spaces.
$f: X \to Y$ is a homotopy equivalence if and only if the homotopy fibers of $f$ are contractible.
I've seen this result mentioned frequently, but I'm struggling to come up with a proof for it.
Sorry for not uploading attempts, but they've all been fruitless.
My spaces are CW complexes and path connected. The point of the homotopy fiber is to turn any map into a fibration, i.e. replace a nasty map $X\rightarrow Y$, into a fibration sequence
$$ F\rightarrow \tilde X\rightarrow Y $$
Now consider the long exact sequence of homotopy groups of a fibration
$$ \ldots \rightarrow \pi_n(F)\rightarrow \pi_n(\tilde X)\rightarrow \pi_n( Y)\rightarrow \pi_{n-1}(F)\rightarrow \ldots $$
All the maps $\pi_n(\tilde X)\rightarrow \pi_n(Y)$ are isomorphisms if and only if $\tilde f$ is a homotopy equivalence (Whitehead's theorem). This occurs precisely if and only if $\pi_n(F)=0$ for all $n$. But this implies that $F$ is contractible.