Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a continuous real function and we consider its graph identified by the subset:


of $\mathbb{R}^{n+1}$ equipped with the usual euclidean topology that induced on $\mathcal{G}(f)$ a subspace topology. Well with these condictions we demonstrate that $\mathcal{G}(f)$ is homeomorphic to $\mathbb{R}^n$. So we consider the function

$$ h:\mathcal{G}(f)\owns(x,f(x))\rightarrow x\in \mathbb{R}^n$$

since the domain of $f$ is $\mathbb{R}^n$, clearly $h$ is surjective on $\mathbb{R}^n$ and then is also injective because if $h(x,f(x))=h(y,f(y))$ then $x=y$; now to prove the assertion we have to demonstrate of that $h$ and $h^{-1}$ are continuous funcion or that $h$ is open and continuous, but unfortunately I'm not be able to do this so can someone help me?

  • 3
    $\begingroup$ You have the tag (general-topology). So you may as well prove your result for $f : X \to Y$ where $X,Y$ are topological spaces. Christoph's proof works just as well. $\endgroup$
    – GEdgar
    Feb 5, 2020 at 16:03

2 Answers 2


If $f: X \to Y$ is continuous, $\Gamma(f)=\{(x,f(x)): x \in X \} \subseteq X \times Y$ is homeomorphic to $X$.

The homeomorphism is the obvious $h: X \to X \times Y$ defined by $h(x)=(x,f(x))$ which is continuous as a map into $X \times Y$ as $\pi_X \circ h = 1_X$ and $\pi_Y \circ h = f$ are both continuous, by the universal mapping property of the product. And by definition $h[X]=\Gamma(f)$.

And the continuous inverse of $h$ is $\pi_X\restriction_{\Gamma(f)}$, which is continuous as the restriction of a continuous map.

  • $\begingroup$ Could you give me some information of the universal mapping property? Unfortunately I only know to factorization of tensor product: pheraps the latter is a special case of the first? $\endgroup$ Feb 5, 2020 at 18:51
  • $\begingroup$ Maybe I can see things this way? Let be $g:Y\rightarrow Z$ a continuous function beetween two topological spaces and we suppose that the function $f:X\rightarrow Y$ beetween the topological spaces $X$ and $Y$ is such that its composition with $g$ is continuous: so is $f$ continuous? If $g\circ f$ is continuous so for any open set $U$ in $Z$ it result that $(g\circ f)^{-1}(U)=f^{-1}(g^{-1}(U))$ is an open set in $X$ so by the continuity of $g$ we can ovserve that the the inverse imagine by $f$ of the open set $g^{-1}(U)$ is an open set: so could I conclude that $f$ is continuous? $\endgroup$ Feb 5, 2020 at 19:12
  • $\begingroup$ @AntonioMariaDiMauro This property that continuity of compositions gives the continuity of the function is a characteristic property of the product topology (as a special case of an "initial topology"); implied by the minimality of the product topology. This is based on quite general (but elementary) theory. $\endgroup$ Feb 5, 2020 at 21:32
  • $\begingroup$ Okay, unfortunately I didn't yet study the product topology: infact in the post I wrote that the topology on $\mathcal{G}(f)$ is the subspace topology so maybe it could be wise to wait to study the next chapter of my text, where indeed is explained the product topology (I don't know why the autors decided to give me this exemple without the definition of product topology), to understead your explanation. Anyway using the subspace topology is it possible to demostrate the continuity of $h$? $\endgroup$ Feb 5, 2020 at 21:48
  • $\begingroup$ @AntonioMariaDiMauro $h$ is clearly continuous: just apply the definitions of the metric on $\Bbb R^{n+1}$ and $\Bbb R^n$ etc. plus continuity of $f$. Work locally. It's more messy than the product topology proof, but still doable. $\endgroup$ Feb 5, 2020 at 21:56

Note that $h$ is a restriction of the projection $\mathbb R^n\times\mathbb R\to\mathbb R^n$, which you can check is continuous. The inverse of $h$ is the map $h^{-1}\colon \mathbb R^n \to \mathbb R^n\times\mathbb R$ given by $x\mapsto(x,f(x))$. Hence $h^{-1} = (\operatorname{id}_{\mathbb R^n}, f)$ and since both $\operatorname{id}_{\mathbb R^n}$ and $f$ are continuous, so is $h^{-1}$.


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