Let $f:X\rightarrow Y$ be bijective continuous and $Y$ is compact (my $X$ and $Y$ are also Hausdorff.)
Is X necessarily compact?
If not, what can be a $\textit{weak}$ extra condition so that this is true?
My motivation:
Assume that for a $\textit{free}$ group action $\rho:X\times GL(n,\mathbb{C})\longrightarrow X$ the graph $\Gamma$ of the map
$\gamma:X\times GL(n,\mathbb{C})\longrightarrow X\times X,~~~(x,g)\mapsto (x,\rho(x,g))$
is $\textit{closed}$.
I want to show that this action is locally proper (i.e. any point of $X$ has a compact $GL(n,\mathbb{C})$-stable neighborhood to which the restriction of $\rho$ is proper.)
Proof: It suffices to show that, for any $x\in X$ and a compact neighborhood $X\supseteq K_x\ni x$ of $x$, the restriction map $\rho|_{{K_x}\times GL(n,\mathbb{C})}$ is proper. One has
$Im\left(\gamma|_{{K_x}\times GL(n,\mathbb{C})}\right)=\Gamma\cap({K_x}\times X).$
Since the action $\rho$ is free, the map
$\gamma|_{{K_x}\times GL(n,\mathbb{C})}: K_x\times GL(n,\mathbb{C}) \longrightarrow \Gamma\cap({K_x}\times X)$
is injective (and thus bijective.)
For a compact $K\subset X$, we want the set
$\left(\rho|_{{K_x}\times GL(n,\mathbb{C})}\right)^{-1}(K)=\left(\gamma|_{{K_x}\times GL(n,\mathbb{C})}\right)^{-1}\left(\Gamma\cap({K_x}\times K)\right)$
to be compact. This would follow from the statement I ask, since $\Gamma\subset X\times X$ is closed.
No, one could have any compact Hausdorff space for $Y$ and then take $X$ to be the same space with the discrete topology (and $f$ to be the identity map).