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Let $X$ and $Y$ be topological spaces and $f:X\to Y$. The graph of $f$ is the subset $G=\{(x,f(x))\}\subseteq X\times Y$. Show that if $f$ is continuous, then $X$ is homeomorphic to the graph of $f$.

My attempt: Let $h:X\to G$ by $h(x)=(x,f(x))$. The fact that $h$ is a bijection is clear. Now since $g:X\to X$ by $g(x)=x$ is continuous and $f:X\to Y$ is continuous by hypothesis, the universal mapping property of products says that $h$ must be continuous. I'm having trouble showing $h^{-1}$ is continuous though... $g^{-1}:X\to X$ is continuous by the same argument as for $g$, but how would I show that $f^{-1}$ is continuous?

Martin Sleziak
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user124910
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    Older question about the same result: http://math.stackexchange.com/questions/429797/if-f-is-continuous-then-g-is-homeomorphic-to-x – Martin Sleziak Feb 27 '14 at 08:32
  • Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Feb 27 '14 at 08:36

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The map $h^{-1}\colon G\to X$ maps $(x,f(x))$ to $x$. Therefore it can be obtained as the restriction of the projection $X\times Y\to X$ to the subspace $G$ of $X\times Y$. Projections are continuous, and so is an inclusion map of a subspace.

Carsten S
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Your basic idea is good, but try the composition $X\to X\times X \to X\times Y$ defined as follows instead $$ x\mapsto (x,x)\mapsto (x, f(x)) $$ Show that in each step you have a homeomorphism.