# Composite function between manifolds

I have this claim left as an exercise in my course:

Let $f:M\to N$ be some function between two smooth manifolds $M$ and $N$ (respectively of dimensions $m$ and $n$). Prove that, if for any smooth function $\mu:N\to\mathbb{R}$, we have that $\mu\circ f$ is smooth, then $f$ is smooth. (Here, I say smooth to say "smooth of class $C^{k}$", $k\geq 1$). Hint: use the chain rule.

Here is how I tried:

Let $(U_{i},\varphi_{i})_{i\in I}$ be some atlas of $M$ and $(V_{j},\psi_{j})_{j\in J}$ some atlas of $N$. Now take $(i,j)\in I\times J$ such that $f^{-1}(V_{j})\cap U_{j}\neq\emptyset$. Proving that $f$ is smooth means that we need to show $$\psi_{j}\circ f\circ\varphi^{-1}_{i}\left.\right\vert_{\varphi_{i}(f^{-1}(V_{j})\cap U_{i})}$$ is of class $C^{k}$ for any $(i,j)\in I\times J$.

We know that every component of $\psi_{j}$ is smooth and then, by hypothesis on $f$, any component of $\psi_{j}\circ f$ is smooth. I posted that on another forum and their conclusion was I can't state that without constructing a function extending $\psi_{j}$ $C^{k}$-continuously on $N$. We can do that by constructing a function \begin{aligned}\alpha_{j}&=1 &\text{on}\,\,V'_{j}\\ &= 0 &\text{out of}\,\,V_{j} \end{aligned} and $\alpha_{i}$ is some $C^{k}$-continuous function on $V_{j}\setminus V'_{j}$ where $V'_{j}$ is some open set of $N$ such that $\overline{V'}_{j}\subset V_{j}$. If this is true until there, my main problem is to prove the existence of such an open set $V'_{j}$.

I suppose the chain rule is there to prove the smoothness of the composition in the first equation above.

• What you seek is called "partitions of unity". See for example math.toronto.edu/mat1300/partitions.10.pdf – AnatolyVorobey Oct 25 '15 at 19:51
• Thank you for the answer, but in my course, "partitions of unity" come after this exercise, so either my professor is not coherent in the construction of his course, or it does not require partition of unity. Is it possible to show that without partition of unity? In any case, thank your for your help (and this useful link). – MoebiusCorzer Oct 25 '15 at 19:53
• Well, sometimes partitions of unity are introduced pretty early, precisely so that such tricks are possible. If it isn't the case in your course, perhaps your professor copied the exercise from a different course/textbook and didn't give it much thought. It does look trivial and it's easy to miss at a first glance that it requries smth like partitions of unity. But I think it does. You can cut some corners b/c you don't need to glue many of them together, but fundamentally you need something like your $\alpha_i$. – AnatolyVorobey Oct 25 '15 at 23:51
• Thank you for the answer. I discussed that with friends and they told me it is not necessary to take such a $V'_{j}$. Here is what they say: we define $\alpha_{j}$ to be $1$ on $V_{j}$, some $C^{k}$-continuous function on a $\text{neighbourhood}\setminus V_{j}$ of each point of the boundary of $V_{j}$ and $0$ outside. I think it is consistant but you seem so sure about partition of unity that maybe we miss something. – MoebiusCorzer Oct 26 '15 at 16:57
• Look at the simplest example: $N=R$, and $V_j=(0,1)$. You want a smooth function defined on all of $R$ that is $1$ on $(0,1)$ but has compact support. That's not trivial! You can't just say "define it to be $0$ outside $[0,1]$ and mumble mumble $C^k$ at neighbourhoods of $0$ and $1$". That makes no sense. You need to use something like the bump function $e^{-1/x}$ and its properties, i.e. develop the machinery of partitions of unity. If that's what you mean, we're on the same page. If you think you're getting your $\alpha_j$ w/o careful use of something like $e^{-1/x}$, I'd like to see how. – AnatolyVorobey Oct 26 '15 at 23:13