Prove that the If $G(f)$ inherits the product topology from $X \times Y$, then $X$ is homeomorphic to $G(f)$ via the map $x \to (x,f(x))$.

Question

Let $X$ and $Y$ be topological spaces and let $f\colon X \to Y$ be a continuous map. Define $G(f)={(x, f(x)| x \in X)} \subset X \times Y$. If $G(f)$ inherits the product topology from $X \times Y$, then $X$ is homeomorphic to $G(f)$ via the map $x \mapsto (x,f(x))$.

Is there any idea where I can start to prove or disprove the above statement?

Thanks in advance

Answer 1

A function $F: Z \to X \times Y$ (for any space $Z$) is continuous (when $X \times Y$ has the product topology) iff $\pi_X \circ F$ and $\pi_Y \circ F$ are both continuous (and $\pi_X, \pi_Y$ are the projection maps, which are continuous).

If we take $Z=X$ and $F(x)=(x,f(x))$ as suggested, then $F$ is continuous by this criterion as $\pi_X \circ F = \textrm{id}_X$ (always continuous) and $\pi_Y \circ F=f$, which is continuous by assumption.

It's clear that by definition $F$ is a bijection between $X$ and $G(f)$ and its inverse map is $\pi_X\restriction_{G(f)}$, which is also continuous (as the restriction of a continuous map to a subspace).

So $F$ is a homeomorphism between $X$ and $G(f)$.