Coincidence of topology defined by $M\to R^{C^{\infty}(M)}$ with $M$ and $M\to R^{C(M)}$ with $M$?

Question

Consider $M$ a smooth manifold. Let $C^{\infty}(M)$ and $C(M)$ be the set of smooth functions and continuous functions over $M$ respectively. Let $R$ be real number.

Consider weak topologies $T_\infty$ and $T_c$ induced by maps $M\to R^{C^{\infty}(M)}$ and $M\to R^{C(M)}$ respectively where infinite product of $R$ is assumed with product topology. The maps are given by evaluation maps as $x\to \prod_{f\in C^{\infty}(M)} f(x)$ or $\prod_{f\in C(M)} f(x)$ respectively.

Clearly any function is smooth is continuous. Hence, $T_c$ is much finer than $T_\infty$.

$\textbf{Q:}$ I am not going to assume $M$ compact as under sup norm, this may allow approximation of $C(M)$ via $C^\infty(M)$ which is dense inside $C(M)$. However, is $T_c=T_\infty$ here? What norm should one use on $C^\infty(M), C(M)$ here?(i.e. I want to induce a norm structure on $R^{C^\infty(M)}$ and $R^{C(M)}$)

This is related to the definition of manifold in Milnor's Characteristic classes.

Answer 1

The two topologies are the same and coincide with the original topology on $M$. There are many ways to prove this. Here's one quick way using a bit of machinery: by the Whitney embedding theorem we can embed $M$ as a closed submanifold of $\mathbb{R}^n$ for some $n$. Then, every closed subset of $\mathbb{R}^n$ is the vanishing set of a smooth function, and thus by restricting to $M$ the same is true of $M$ (alternatively, instead of using the Whitney embedding theorem, the proof for $\mathbb{R}^n$ can be adapted to work for a general manifold). This means that $T_\infty$ contains the original topology. Since $T_\infty$ is contained in $T_c$ and $T_c$ is contained in the original topology, this means all three topologies are the same.