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Let $f:X\rightarrow Y$ be a continuous function and $G_f$ be its graph. I want to prove that the function $$h:X\rightarrow G_f \\ x\mapsto (x,f(x))$$ is an homeomorphism.

I am aware of the question Proving a homeomorphism when graph of function has product topology. The answer given there did not answer my particular doubts.

My attempt:
Each component of $h$ is continuous, so $h$ is continuous.
Let $x,y\in X$ with $h(x)=h(y)$. Then, $(x,f(x))=(y,f(y))$. In particular, $x=y$, so $h$ is injective.
Let $(x,y)\in G_f$. Then, $y=f(x)$. If we take the point $x$ of $X$ we get $h(x)=(x,y)$, so $h$ is surjective.
We now have to establish the continuity of $h^{-1}$. Let $U$ be an open set of $X$. I want to prove that $(h^{-1})^{\dashv}(U)$ is an open set of $G_f$ (i.e. that there is an open set $A$ of $X\times Y$ such that $(h^{-1})^{\dashv}(U)=A\cap G_f$).

Questions:
$\bullet$ How to finish the proof?
$\bullet$ I've read that it suffices to show that $h$ is open, but I have to idea why.

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    Do you have further informations on the spaces X and Y? – 2017-02-04
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    If you write $(h^{-1})^{-1}(U) = \{ (x,f(x)) : \text{ some condition involving } x\}$, it is easy to see what $A$ to use. – 2017-02-04
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    @Maczinga No, I don't. – 2017-02-04
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    The inverse function is the restriction to $G_f$ of the projection $X\times Y\to X$, so it is obviously continuous. – 2017-02-04
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    Note that $h^{-1}=\pi_1\restriction_{ G_f}$ – 2017-02-04
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    For a bijection $h: A \rightarrow B$, $(h^{-1})^{-1}[U] = \{b \in B: h^{-1}(b) \in U\} = h[U]$ for all $U$ (also open or closed ones) subset of $A$. So $h$ open or closed is equivalent to $h^{-1}$ continuous. – 2017-02-04
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    Continuity of $h$ itself is clear from the universal criterium for cotinuity into a product space: $\pi_1 \circ h = id_X$, $\pi_2 \circ h = f$, and both are continuous. – 2017-02-04

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$G_f$ is a subspace of $X \times Y$ and so its topology is the subspace topology of the quotient topology. If $U \subset X$ is open, then $U \times Y$ is open in $X \times Y$, and so $(f^{-1})^{-1}(U) = f(U) = (U \times Y) \cap G_f$ is open in $G_f$.

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    You used $(f^{-1})^\dashv=f(U)$. When exactly is this valid? When $f$ is bijective? – 2017-02-04
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    Bijective is certainly sufficient. You don't need anything else for this problem because your map $f : X \to G_f$ is a bijection as you showed. If one does not have bijectivity, this holds if and only if $U \subset X$ is a "saturated subset" with respect to $f$, meaning that for all $x,y \in X$ if $f(x)=f(y)$ and $x \in U$ then $y \in U$. – 2017-02-04