When is it true that $F: N \rightarrow M_1 \times \cdots \times M_k$ is smooth iff each component map is smooth?

438 Views Asked by At

I'm reading Introduction to Smooth Manifolds by John M. Lee, and I have a question about Proposition 2.12 (=Problem 2-2) on P.36.

Proposition 2.12. Suppose $M_1, \cdots, M_k$ and $N$ are smooth manifolds with or without boundary, such that at most one of $M_1, \cdots, M_k$ has nonempty boundary. For each $i$, let $\pi_i: M_1 \times \cdots \times M_k \rightarrow M_i$ denote the projection onto the $M_i$ factor. A map $F: N \rightarrow M_1 \times \cdots \times M_k$ is smooth if and only if each of the component maps $F_i = \pi_i \circ F: N \rightarrow M_i$ is smooth.

The main idea of my solution is:

  • Suppose $F$ is smooth. It is not hard to prove that each $\pi_i$ is smooth. By Proposition 2.10(d), the composition of smooth maps is smooth. Thus each $F_i$ is smooth.
  • Suppose each $F_i$ is smooth. Let $p \in N$ be given. Then for each $i$, there exist charts $(U_i, \phi_i), (V_i, \psi_i)$ such that $p \in U_i, F_i(U_i) \subset V_i$ and $\psi_i \circ F_i \circ \phi_i^{-1}$ is smooth. Let $U = \cap U_i$. Then $(\phi_1\vert_U, U)$ and $(\psi_1 \times \cdots \times \psi_k, V_1 \times \cdots V_k)$ are charts that satisfy the properties in the definition of a smooth map.

The problem with my solution is that it does not use the condition that at most one of $M_1, \cdots, M_k$ has nonempty boundary. Which part of my proof would fail without that condition?

Thank you!

1

There are 1 best solutions below

0
On BEST ANSWER

This question is actually answered in an earlier chapter in the textbook.

The first paragraph of P.29 of Chapter 1 mentions that "a finite product of smooth manifolds with boundary cannot generally be considered as a smooth manifold with boundary" with some explanation.

So, the condition was necessary to make sure that $M_1 \times \cdots \times M_k$ is a smooth manifold with boundary, but it was not a condition that I had to use directly in the solution.