Proposition

Let $(X_1, \mathcal{T}_1)$ and $(X_2, \mathcal{T}_2)$ be topological spaces and let $f: X_1 \rightarrow X_2$ be a bijective map. Show that $f$ is a homeomorphism if and only if $f(\mathcal{T}_1) = \mathcal{T}_2$ in the sense that $U \in \mathcal{T}_1$ if and only if $f(U) \in \mathcal{T}_2$.

Solution

Suppose $f$ is a homeomorphism. Since $f$ is open, $f(U) \in \mathcal{T}_2$ for any $U \in \mathcal{T}_1$. Let $U \subset X_1$. Suppose $f(U) \in \mathcal{T}_2$. Since $f$ is continuous, $f^{-1}(f(U)) \in \mathcal{T}_1$. Since $f$ is bijective, $f^{-1}(f(U)) = U$. Thus $U \in \mathcal{T}_1$.

Suppose $U \in \mathcal{T}_1$ if and only if $f(U) \in \mathcal{T}_2$. Then $f$ is open because $f(U)$ is open for each open $U$. Let $U \in \mathcal{T}_2$. Then $f(f^{-1}(U)) = U$ since $f$ is bijective. Since $f(f^{-1}(U))$ is open in $\mathcal{T}_2$, $f^{-1}(U)$ is open in $\mathcal{T}_1$. Thus $f$ is continuous. Therefore, $f$ is a homeomorphism.