Topological embedding iff $X$ homeomorphic to $f(X)$

Question

I want to prove that $f: X \to Y$ is an embedding iff it induces a homeo $\tilde f: X \to f(X)$.

Recall that an injective homeo is an embedding if $\tau_X = \{f^{-1}(V): V \in \tau_Y \}$.

Now, if $f$ is an embedding, because for all $V$ open in $Y$ we have $f^{-1}(V) = \tilde f^{-1}(V \cap f(X))$, $\tilde f$ is a homeo.

Conversely, suppose we have a homeo $\tilde f : X \to f(X)$. We want to prove that $\tau_X = \{f^{-1}(V): V \in \tau_Y \}$. Clearly, since $f$ is continuous, $f^{-1}(V) \in \tau_X$ for all $Y \in \tau_Y$. How can I prove these are the only opens? I guess I should also use $f$ injective somewhere.

Thanks.

Answer 1

Take any open set $U\subset X$. Since $\tilde f$ is a homeomorphism, the image $\tilde f(U)\subset Y$ is open in $f(X)$. A homeomorphism always maps an open set to an open set, as can be seen by using the definition of continuity for the inverse $\tilde f^{-1}$. By the definition of the relative topology of $f(X)\subset Y$, there is an open set $V\subset Y$ so that $\tilde f(U)=f(X)\cap V$.

Now: $$f^{-1}(V)=\tilde f^{-1}(V)=\tilde f^{-1}(\tilde f(U))=U$$ so $U$ is of the desired form.

Answer 2

The fact that $\tilde f:X\to f(X)$ is a homeomorphism tells us that there is a one-to-one correspondence between sets open in $X$ and sets open in $f(X)$.

More precisely the maps $\phi:\tau_X\to\tau_{f(X)}$ and $\theta:\tau_{f(X)}\to\tau_X$ prescribed by $U\mapsto f(U)$ and $V\mapsto f^{-1}(V)$ respectively are inverses of each other.

So for $U\in\tau_X$ we find: $$U=\theta\phi(U)=f^{-1}(f(U))$$where $f(U)\in\tau_{f(X)}$.

This means that $f(U)=f(X)\cap V$ for some $V\in\tau_Y$ so that $U=f^{-1}(f(X)\cap V)=f^{-1}(V)$.