Graphs of Continuous Functions are Topological Manifolds

Last updated Nov 1, 2022

Let $M, N$ be Topological Manifolds of Manifold Dimensions $m,n$ respectively and let $f: M \to N$ be a Continuous Function. Then the graph of $f$, $\Gamma(F) \subset M \times N$, with the Subspace Topology is a Topological Manifold of Manifold Dimension $n$.

In fact, $\Gamma(f)$ is a Topological Submanifold of $M \times N$ endowed with the product manifold.

We simply check the definition of a Topological Manifold.

1. Subspace Topology inherits Second Countability
2. Subspace Topology inherits Hausdorfness
3. Consider the Projection Map $\pi_{1} : \Gamma(f) \to M$ that sends $(x, y) \mapsto x$. First we show $\pi_{1}$ is a Homeomorphism. Indeed $\pi_{1}$ is continuous. Observe that $\rho : M \to \Gamma(f)$ that sends $x \mapsto (x, f(x))$ is the Function Inverse of $\pi_{1}$ since $\pi_{1} \circ \rho$ sends $x \mapsto (x, f(x)) \mapsto x$ and $\rho \circ \pi_{1}$ sends $(x, f(x)) \mapsto x \mapsto (x, f(x))$. Furthermore $\rho$ is continuous. To see this, note that because the Subspace Topology preserves Bases, $$\mathcal{B} := {(U_{1} \times U_{2}) \cap \Gamma(f) : U_{1} \subset M \text{ is open}, U_{2} \subset N \text{ is open}}$$ is a Topological Basis for $\Gamma(f)$. For $V \in \mathcal{B}$, where $V = (U_{1} \times U_{2}) \cap \Gamma(f)$ for $U_{1}, U_{2}$ Open in $M,N$ respectively, $$\rho^{-1}(V) = \rho^{-1}((U_{1} \times U_{2}) \cap \Gamma(f)) = f^{-1}(U_{2}) \cap U_{1}.$$ The last equality follows because $$\begin{align*} x \in f^{-1}(U_{2}) \cap U_{1} &\Leftrightarrow x \in U_{1} \wedge f(x) \in U_{2}\\ &\Leftrightarrow (x, f(x)) \in (U_{1} \times U_{2}) \cap \Gamma(f)\\ &\Leftrightarrow \rho(x) \in (U_{1} \times U_{2}) \cap \Gamma(f)\\ &\Leftrightarrow x \in \rho^{-1}((U_{1} \times U_{2}) \cap \Gamma(f)). \end{align*}$$So $\rho$ is a Continuous Function because A Function is Continuous iff it takes Basis Sets back to Open Sets and $f^{-1}(U_{2}) \cap U_{1}$ is Open ($f$ is continuous). Thus $\pi_{1}$ is a Homeomorphism. Now let $(x, f(x)) =: p \in \Gamma(f)$ and let $(U, \varphi)$ be a Coordinate Chart containing $x$ for $M$. Then because $\pi_{1}$ is a Homeomorphism and Composition of Homemorphisms is a Homeomorphism, $\varphi \circ \pi_{1}$ us a Homeomorphism from $U \times f(U) = (U \times N) \cap \Gamma(f)$ to $\varphi(U) \subset \mathbb{R}^{n}$. Since $p \in \Gamma(f)$ was arbitrary, these Coordinate Charts cover $\Gamma(f)$.

$\blacksquare$