Continuity, homeomorphism, topology

Question

Let $f: X \rightarrow Y$ be a function we define the graph of $f$ as $Γf = \{x × f (x): x \in X\} \subset X × Y$. Show that when $f$ is continuous it is true that $Γf$ is homeomorphic to X. Why the previous theorem shows that the surface defined by the equation $z = xy$ is homeomorphic to $\mathbb{R}^2$? could someone help me with this problem? It is not clear to me how the going and then the return goes

Answer 1

There are not many ways to find a homeomorphism between $X$ and $\Gamma f$. The simplest idea i can think of is : $$\Phi : \begin{align*}X &\to \Gamma f\\x &\mapsto (x, f(x)).\end{align*}$$

$\Phi$ is continuous because $f$ is ($\Gamma f$ having the induced topology from the product topology on $X \times Y$). It is trivially bijective, and its inverse is : $$\Phi^{-1} : \begin{align*}\Gamma f &\to X\\(x, y) &\mapsto x.\end{align*}$$ This map is continuous since it is just a restriction of the projection $X \times Y \twoheadrightarrow X$.