Homeomorphism to graph

general-topologymanifoldsproduct-spacesolution-verification

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.

Please tell me if this is correct first and then include additional comments, please.

Best Answer

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.

The corollary is also fine; presumably your definition of manifold includes that it must be second countable and Hausdorff. Of course, local Euclideanness plus these two properties are preserved by homeomorphisms...

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