# Continuous functions and homeomorphisms

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• Jun 12th 2009, 08:45 AM
Inti
Continuous functions and homeomorphisms
Let $\displaystyle X$ and $\displaystyle Y$ be topological spaces and $\displaystyle f : X \longrightarrow Y$ a function. The graph of $\displaystyle f$ is the set $\displaystyle G_f = \{(x,f(x)) : x \in X\}$.

Prove that $\displaystyle f$ is continuous if and only if the function $\displaystyle \Phi : X \longrightarrow G_f$ given by $\displaystyle \Phi(x) = (x,f(x))$ is a homeomorphism where $\displaystyle G_f$ has the topology relative to the product topology $\displaystyle X \times Y$.
• Jun 13th 2009, 09:09 AM
aliceinwonderland
Quote:

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

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

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

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

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

Then g is continuous if and only if the functions

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

are continuous.

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

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

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

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

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

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

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

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

Note: This is my attemp to this problem. Any feedback or correction is welcomed.
• Jun 15th 2009, 05:51 AM
Inti
Cool, it seems to be ok to me.
Thank you very much!!!