Continuous Bijection of Top Spaces It is well known that a continuous bijection of compact hausdorff topological spaces is a homoemorphism. I am wondering, is it true that a continuous bijection of compactly generated spaces is a homemorphism?
I am also interested in other generealizations of this theorem.
Thanks so much!

Here is a negative result that precludes many generalizations you might consider (though it doesn't quite address your case where you require both the domain and the codomain to be compactly generated).
Theorem: Let $X$ be a regular Hausdorff space such that every continuous bijection $X\to Y$ to a Hausdorff space $Y$ is a homeomorphism.  Then $X$ is compact.
Proof: Suppose $X$ is not compact; let $\mathcal{A}$ be a collection of nonempty closed subsets of $X$ which is closed under finite intersections but such that $\bigcap \mathcal{A}=\emptyset$.  Fix a point $x\in X$ and let $Y$ be $X$ with the topology such that a set is $U$ is open iff it is open in $Y$ and if $x\in U$, then $U$ contains an element of $\mathcal{A}$.  Clearly the identity map $X\to Y$ is a continuous bijection.  Since $x\not\in\bigcap \mathcal{A}$, there is some $A\in\mathcal{A}$ such that $x\not\in A$, and then $X\setminus A$ is open in $X$ but not in $Y$, so the identity map is not a homeomorphism.
It remains to be checked that $Y$ is Hausdorff.  For points in $Y$ besides $x$ this is trivial, so we must just show that if $y\in Y$ is distinct from $x$ then $x$ and $y$ are separated by open sets.  Since $y\not\in\bigcap \mathcal{A}$, there is some $A\in\mathcal{A}$ such that $y\not\in A$.  By regularity, we can then separate $y$ and $A\cup\{x\}$ by open sets of $X$, and these will still be open in $Y$ since the one that contains $x$ also contains $A$.
