If $f$ is continuous then $G$ is homeomorphic to $X$.

Let $f: X \to Y$ be a function. The graph of $f$ is defined to be the set $G = \{(x, f(x)) : x \in X\}$. Prove that if $f$ is continuous then $G$ is homeomorphic to $X$.

an anyone suggest me the process of solving this problem, thanks.

Solution 1:

HINT: There is only one reasonable candidate for a homeomorphism from $X$ onto $G$, namely, the map

$$F:X\to G:x\mapsto\langle x,f(x)\rangle\;;$$

use the definitions of continuity and the product topology to show that $F$ is a homeomorphism. It’s very easy to show that $F$ is a continuous bijection. Since it’s a bijection, it at least has an inverse $F^{-1}$. Show that $F^{-1}=\pi_1\upharpoonright G$, where $\pi_1:X\times Y:\langle x,y\rangle\mapsto x$ is the projection map to the first factor. What do you know about projection maps?

Solution 2:

Hints: Let $\varphi : x \mapsto (x,f(x))$.

• Show that $\varphi$ is a continuous bijection from $X$ to $G$.
• Show that the projection on the first coordinate $\text{pr}_1 : X \times Y \to X$ is continuous and $\text{pr}_1 \circ \varphi =\operatorname{Id}_X$.