Continuous functions and homeomorphisms

**Inti** · June 12th 2009, 08:45 AM

Let $X$ and $Y$ be topological spaces and $f : X \longrightarrow Y$ a function. The graph of $f$ is the set $G_f = \{(x,f(x)) : x \in X\}$ .

Prove that $f$ is continuous if and only if the function $\Phi : X \longrightarrow G_f$ given by $\Phi(x) = (x,f(x))$ is a homeomorphism where $G_f$ has the topology relative to the product topology $X \times Y$ .

**aliceinwonderland** · June 13th 2009, 09:09 AM

Originally Posted by Inti

Let $X$ and $Y$ be topological spaces and $f : X \longrightarrow Y$ a function. The graph of $f$ is the set $G_f = \{(x,f(x)) : x \in X\}$ .

Prove that $f$ is continuous if and only if the function $\Phi : X \longrightarrow G_f$ given by $\Phi(x) = (x,f(x))$ is a homeomorphism where $G_f$ has the topology relative to the product topology $X \times Y$ .

Definition. Let X and Y be a topological space; let $h : X \rightarrow Y$ be a bijection.
If both the function h and the inverse function $h^{-1}:Y \rightarrow X$ are continuous, then h is called a homeomorphism.

Lemma 1. Let $g:A \rightarrow X \times Y$ be given by the equation

$g(a) = (g_1(a), g_2(a))$ .

Then g is continuous if and only if the functions

$g_1:A \rightarrow X$ and $g_2:A \rightarrow Y$

are continuous.

( $\Rightarrow$ )
Suppose f is a continuous function.
Since f is a well-defined function, it is clear that $\Phi : X \longrightarrow G_f$ defined by $x \mapsto (x, f(x))$ is a bijection. Let $U \times V$ be an open set of $G_f$ relative to the product topology $X \times Y$ containing an arbitrary point $(x_1, f(x_1)) \in G_f$ . It follows that

$\Phi^{-1}(U \times V) = U \cap f^{-1}(V)$ .

Since both an identity map and f are continuous functions, $U$ and $f^{-1}(V)$ are open, so is their intersection. It follows that $\Phi^{-1}(U \times V)$ is open. Thus, $\Phi$ is continuous.

Now, we need to show that $\Phi ^{-1} : G_f \longrightarrow X$ defined by $(x, f(x)) \mapsto x$ is a continuous function. Let U be an open set containing an arbitrary point of $x_1 \in X$ . Then, $(\Phi ^{-1})^{-1}(U) = K$ , where $K$ is an intersection of a neighborhood of $(x_1, f(x_1)) \in U \times V \subset X \times Y$ with $G_f$ . Specifically,

$K = (U \times M) \cap G_f = (U \times V) \cap G_f$ ,

where M is a set $f(U) \subset Y$ and V is an open set in Y containing M.

Since $G_f$ has the topology relative to the product topology $X \times Y$ , $K$ is open in $G_f$ . Thus, $\Phi ^{-1}$ is continuous.
By the defintion given the above, $\Phi$ is a homeomorphism.

( $\Leftarrow$ )
Suppose $\Phi$ is a homeomorphism.
Since an identity map is continuous and $\Phi : X \longrightarrow G_f$ given by $\Phi(x) = (x,f(x))$ is continuous by hypothesis, $f$ is continuous by lemma 1.

Note: This is my attemp to this problem. Any feedback or correction is welcomed.

**Inti** · June 15th 2009, 05:51 AM

Cool, it seems to be ok to me.
Thank you very much!!!

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