I have the following question: Let $f: X \rightarrow Y$ be a open, continuous map and let $X' \subseteq X$ be a closed subspace such that the restriction $f|_{X'}: X' \rightarrow f(X')$ is a bijection. Does this already imply that $f|_{X'}$ is proper?
My attempt of a proof looks like this: If $C\subseteq f(X')$ is compact and $(U_i \cap X')_{i\in I}\subseteq X'$ is an open cover of $f|_{X'}^{-1}(C)$, the family $(U_i \cup (X')^c)_{i\in I} \subseteq X$ is an open cover of $f^{-1}(C)$. Because $f$ is open $(f(U_i \cup (X')^c))_{i\in I} \subseteq Y$ is an open cover of $C$ in $Y$, so $(f(U_i \cup (X')^c) \cap f(X'))_{i\in I} \subseteq Y$ is an open cover of $C$ in $f(X')$. Now we can use compactness of $C$ to find a finite subcover. But I don't see how to proceed from here. Is it even possible? Can anyone help?
The claim is false. Consider $X=\Bbb R$, $Y=S^1$, $X'=[0,\infty)$ and $f(x)=(\cos(4\arctan x),\sin(4\arctan x))$. $f(X')=S^1$ but $\left.f\right\rvert_{X'}^{-1}[S^1]=[0,\infty)$.