When is a continuous bijective map a homotopy equivalence? When is a continuous bijective map a homotopy equivalence? 
If the spaces are compact Hausdorff then a continuous bijective map is a homeomorphism (and so a homotopy equivalence). Are there weaker conditions that ensure it’s at least a homotopy equivalence? Or even just a weak homotopy equivalence?
 A: This is not an answer, but it is too long for a comment:
First of all, to deduce that a continuous bijective map $f \colon X \to Y$ is a homeomorphism it is sufficient for $X$ to be compact and $Y$ to be Hausdorff (there is no need for both to be compact and Hausdorff at the same time). 
I don't know whether there are reasonable conditions for a fixed continuous bijection $f \colon X \to Y$ such that the map becomes a homotopy equivalence (apart from conditions that tell you that $f$ is a homeomorphism, which then implies that $f$ is a homotopy equivalence...)
But maybe it is interesting to see, that there are easy examples for spaces that are not homeomorphic, but homotopy equivalent:
Consider the punctured plane, i.e. $\mathbb{R}^2 \setminus\{0\}$. This space is homotopy equivalent to $S^1 \subset \mathbb{R}^2$ (exercise!). But the former is $\textbf{not}$ compact whereas the latter is compact. In particular there does not exist a homeomorphism between $S^1$ and $\mathbb{R}^2\setminus\{0\}$.
A: There are no general sufficient conditions for this to hold, besides some trivialities, e.g. when the continuous bijections is a homeomorphism or when the spaces are contractible. See also here. 
I know, however, one nontrivial theorem, due to Dowker:
C.H. Dowker, Topology of metric complexes. 
Amer. J. Math. 74, (1952) 555–577. 
He proves that the identity map that is a continuous bijection 
$id: (X,d)\to X$ from a metric simplicial complex to itself (where the target is equipped with the CW complex topology) while (in general) is not a homeomorphism, is, nevertheless, a homotopy equivalence.  
