The graph of a continuous function is a topological manifold

Question

Let $ U$ be an open set in $\mathbb{R}^n$, and let $f: U \to \mathbb{R}^k$ be a continuous map. The graph of the function $f $, denoted $\Gamma(f)$, is defined as $ \Gamma(f) = \{(x, y) : x \in U, y = f(x)\}.$

Prove that $\Gamma(f) $ is an $n $-dimensional topological manifold.

Define the projection map $\pi_1: \mathbb{R}^n \times \mathbb{R}^k \to \mathbb{R}^n $, and let $\varphi $ be the restriction of $ \pi_1 $ to $ \Gamma(f) $.

I already know that $ \Gamma(f) $ is a Hausdorff space and is second-countable. However, regarding the third condition in the definition of a topological manifold, I don't fully understand how $ \varphi $ can be homeomorphic to an open subset of $ \mathbb{R}^n $. Below is my attempt.

Since $ \pi_1 $, as a projection map, is continuous, its restriction to $ \Gamma(f) $, i.e., $ \varphi $, is also continuous. Now I attempt to prove that its inverse is also continuous. Since $ \Gamma(f) $ is a subspace of $ \mathbb{R}^n \times \mathbb{R}^k $, each open set $ O $ in $ \Gamma(f) $ has the following form: $ O = (S \times T) \cap \Gamma(f) = (S \times T) \cap \{(x, y) : x \in U, y = f(x)\}, $ where $ S $ is an open set in $ \mathbb{R}^n $ and $ T $ is an open set in $ \mathbb{R}^k $. Hence, $ \varphi(O) = S \cap U, $ and by the definition of open sets in a topological space, $ S \cap U $ is an open set.

Is there any problem with my proof, or is there anything that needs improvement？

Answer 1

I didn't check your proof carefully (looks right though), but I'd proffer that it's not needed. The inverse to $\phi$ is the map $x \mapsto (x, f(x))$, which is continuous since a map into a product is continuous iff its component functions are, and $x \mapsto x$ and $x \mapsto f(x)$ are both continuous.

Answer 2

Your proof is correct, but it can be done easier with the following general theorem:

Let $f : X \to Y$ be a continuous map between topological spaces $X,Y$ and $\Gamma(f) = \{(x,f(x)) \in X \times Y \mid x \in X\}$ (with the subspace topology inherited from $X \times Y)$. Then $i : X \to \Gamma(f), i(x) = (x, f(x))$, is a homeomorphism whose inverse is the restriction $p = \pi_X \mid_{\Gamma(f)}: \Gamma(f) \to X$ of the projection $\pi_X : X \times Y \to X$.

This is fairly obvious. Clearly $i$ and $p$ are continuous with $p \circ i = id$ and $i \circ p = id$.

Thus in your question $\Gamma(f)$ is homeomorphic to $U$ which is an $n$-dimensional topological manifold, and therefore $\Gamma(f)$ is itself an $n$-dimensional topological manifold.