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I am asked to prove the Whitney's approximation theorem:

Any continuous map $F\colon M \rightarrow N$ is continuously homotopic to a smooth map.

By somehow using this convolution: $$ (\xi \ast h_t)(x) = \int_\mathbb{R} h_t(x-y)\xi(y)dy $$ where: \begin{align} h_t(x) &\colon = \frac{g_t(x)}{\int_{-1/t}^{1/t}g_t(s)ds}\\\\ g_t(x) &\colon = f(1+t^{-1}x) f(1-t^{-1} x) \text{ for } t>0\\\\ f(t) &\colon = \begin{cases} e^{-1/t} &\text{ if } t>0\\\\ 0 &\text{ if } t\leq 0 \end{cases} \end{align}

The convolution has good properties, i.e.:

  1. Given $\xi\colon\mathbb{R}\rightarrow \mathbb{R}$ continuous, $\xi\ast h_t$ is smooth for every $t$.
  2. $\xi\ast h_t \to \xi$ as $t \to 0$ in the $C^0$ norm.

MY TRY

My objective is to somehow define $F$ in terms of $\mathbb{R}\rightarrow \mathbb{R}$ continuous maps. Then I would be able to convolute them by $h_t$.

Given a point $p\in M$, there exists a chart $(U,\varphi\colon U\rightarrow \mathbb{R}^m)$ s.t. $p\in U$. Define $\tilde{p}\colon = \varphi(p)$. Then: $$ F(p) = (F\circ \varphi^{-1})(\tilde{p}) = (F\circ \varphi^{-1})(\tilde{p}_1,\dots,\tilde{p}_m) $$

Now we have maps $F\circ \varphi^{-1}$ that go from an open set $\varphi(U)$ of $\mathbb{R}^m$ to $N$.

Let's proceed similarly for $N$. There exists a chart $(V,\psi\colon N\rightarrow \mathbb{R}^n)$ in $N$ s.t. $F(p)\in V$. Let me define the open set $V'\colon=V\cap F(U)$. Then the following map makes sense: $$ \psi^{-1}\circ \psi \circ F \circ \varphi^{-1}\colon \varphi(U) \longrightarrow V' $$ and it coincides with $F$ where it is defined. So, following the previous: $$ F(p) = (\psi^{-1}\circ \psi \circ F \circ \varphi^{-1})(\tilde{p}_1,\dots,\tilde{p}_m) = \psi^{-1}((\psi \circ F \circ \varphi^{-1})(\tilde{p}_1,\dots,\tilde{p}_m)) $$

Now, the map $\psi \circ F \circ \varphi^{-1}$ goes from $\varphi(U)\subset \mathbb{R}^m$ to $\psi(V')\subset \mathbb{R}^n$. And I'm lost right here.

I have a feeling that we can "split" the continuous map $\psi \circ F \circ \varphi^{-1}$ in terms of continuous maps from $\mathbb{R}$ to $\mathbb{R}$, but I don't know how.

Any help will be appreciated. Thanks a lot!

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    $\begingroup$ Do you have a compactness assumption on, say, $N$? If $N$ is embedded it becomes quite easier. $\endgroup$
    – Aphelli
    Commented Nov 21, 2019 at 11:49
  • $\begingroup$ @Mindlack The only assumption is that both $M$ and $N$ are smooth manifolds... $\endgroup$
    – Txordi
    Commented Nov 21, 2019 at 18:58

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