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Let $X,d$ be a metric space .Let $f:X\to \mathbb R$ be a continuous function.Define $G(f)=\{(x,f(x)):x\in X\}$.

Prove that $G(f)$ is homeomorphic to $X$.

My try:

Since $f$ is continuous then $G(f)$ is a closed set. To define a homeomorphism we need a mapping $g:G(f)\to X$ .Can $g$ be defined as $g((x,f(x))=x$?i.e the projection map.

To check $g$ is a homeomorphism:

Let $U$ be closed in $X$.to show $g^{-1}(U)$ is closed in $G(f)$.Now $g^{-1}(U)=U\times f(U)$ how to show it is closed in $G(f)$??

Next to show that $g$ is a closed map.How to proceed here?

It will be great if someone could check if it is so.

Learnmore
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2 Answers2

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Show that a continuous application is homeomorphic to his domain.

Let $f:M\to N$ continuous. The graphic of $f$ is the subset $G(f)$ of the cartesian product $M\times N$, define by $G(f)=\{(x,f(x));x\in M\}$. The application $\bar f:M \to G(f)\subset M\times N$, with $\bar f(x)=(x,f(x))$, is continuous since his coordinates are continuous.

The inverse $G(f)\to M$, defined by $(x,f(x))\to x$ is continuos, since is equal to restriccion $p_{1}\vert G(f)$ of the projection $p_{1}:M\times N\to M$. Therefore, $\bar f:M\to G(f)$ is a homeomorphism.

pablocn_
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HINT: Show, if you’ve not already done so, that the projection map $\pi:X\times\Bbb R\to X$ is continuous and open. (This is true of all projection maps.) Its restriction to $G(f)$ is then easily seen to be continuous and open, and it’s not hard to check that this restriction is a bijection and hence a homeomorphism.

Brian M. Scott
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  • restriction of a cont map is cont but will rest of an open map be open?I dont find it true – Learnmore May 22 '15 at 16:19
  • @learnmore: It's not true in general, but it is true in this specific setting. – Brian M. Scott May 22 '15 at 16:37
  • Suppose $U\times V$ be a basic open set in $X\times \mathbb R$ then $U$ is open in $X$ then $\pi(U\times V)=U$ is open in X;is it right?? – Learnmore May 22 '15 at 17:27
  • where are we using that $f$ is cont? – Learnmore May 22 '15 at 17:28
  • @learnmore: Continuity of $f$ is used to show that the restriction of the projection to $G(f)$ is open. Alternatively, and possibly easier, use it to show that the inverse of the restriction of the projection to $G(f)$, which is the map $x\mapsto \langle x,f(x)\rangle$, is continuous. – Brian M. Scott May 22 '15 at 18:58