# Homeomorphism to graph

Suppose $$f:X\rightarrow Y$$ is a continuous map. Define the $$\textbf{graph of f}$$ to be $$\Gamma(f)=$$ $$\{$$ $$(x,y)$$ $$\in X\times Y$$ $$:$$ $$y=f(x)$$ $$\}$$ . Then the map $$\phi_f: X \rightarrow \Gamma(f)$$, given by $$\phi_f(x)=(x,f(x))$$ is a homeomorphism.

Solution:

Observe that $$\phi_f$$ is bijective, because it has an inverse, $$\phi^{-1}_f : \Gamma(f) \rightarrow X$$ $$,$$ $$\phi^{-1}_f(x,f(x))=x$$. Hence $$\phi_f(X)=\Gamma(f)$$. I will consider the function $$\phi_f$$ from $$X$$ to $$X \times Y$$ and show that it is continuous. Since $$\pi_1 \circ \phi_f$$ $$=$$ $$id_X$$ and the identity is continuous, this particular composition is continuous. Since $$\pi_2 \circ \phi_f$$ $$=$$ $$f$$ and $$f$$ is continuous by assumption, it follows that this composition is continuous and that $$\phi_f : X\rightarrow X\times Y$$ is continuous. Restricting the codomain to $$\Gamma(f)$$ implies that $$\phi_f : X\rightarrow \Gamma(f)$$ is continuous. Since it is clearly surjective, it follows that it is surjective topological embedding, which is a homeomorphism.

Corollary: Let $$U\subseteq \mathbb{R}^n$$ be an open subset and $$f: U\rightarrow \mathbb{R}^k$$ be any continuous function. Then $$\Gamma(f) \subseteq \mathbb{R}^{n+k}$$ is a manifold.

Proof: Clearly the graph of $$f$$ is second countable and hausdorff as a subspace. Let $$p \in \Gamma(f)$$. Then by the above theorem, a homeomorphism between $$U$$ and $$\Gamma(f)$$ exists. Hence the graph of $$f$$ is manifold.

The proof that $$\phi_f$$ is continuous is fine, using the two compositions with projections. The inverse is even simpler: that's just the restriction of the continuous first projection to $$\Gamma(f)$$ so easily continuous. So right away it's a homeomorphism between $$X$$ and $$\Gamma(f)$$: $$\Gamma$$ is a bijection that is continuous and has a continuous inverse too.