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Let $M$ , $N$, and $P$ be smooth manifolds with or without boundary.

Every constant map $c: M\rightarrow N$ is smooth.

Proof: Let $c: M \rightarrow N$ be a constant map. Let $p \in M$. Smoothness of $c$ means there are charts $(U,\phi)$ of $p$ and $(V,\psi)$ of $c(p)$ such that $c(U) \subseteq V$ and $\psi \circ c \ \circ \phi^{-1} $ is smooth. Since $c$ is a constant map we know that $c(p)=y$ for every $p \in M$.

This is as far as I got with the proof. I'm a bit lost on how to finish the proof using the fact that c is a constant map to show that $c: M \rightarrow N$ is smooth.

I'd appreciate hints or advice instead of a full solution to the problem that way it doesn't spoil the problem for me.

I'm using Lee's Introduction to Smooth Manifolds.

Alexander King
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  • I think you meant $\phi \circ c \circ \psi^-$ is smooth, right? Because otherwise your composition doesn't make sense. –  Jun 22 '17 at 18:50
  • Are you allowed to use the fact that constant maps between open subsets of Euclidean spaces are smooth? (If not, this is easy to prove) – M10687 Jun 22 '17 at 18:51

3 Answers3

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We have chosen some chart $\phi \colon U \rightarrow \phi(U) \subseteq \mathbb{R}^n$ on $M$ and a chart $\psi \colon V \rightarrow \psi(V) \subseteq \mathbb{R}^m$ on $N$. We want to show that $\psi \circ c \circ \phi^{-1}$ is smooth given $c$ is constant.

Hints: 1. What is the domain and codomain of the map $\psi \circ c \circ \phi^{-1}$

  1. Is a constant map $k \colon \mathbb{R}^n \rightarrow \mathbb{R}^m$ smooth?
Radek Suchánek
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  • Domain would be R^n and codomain would be R^m. wouldn't the dimensions need to be the same for k: R^n ---> R^m? – Alexander King Jun 22 '17 at 19:40
  • The dimension to be the same is unnecessary. The question is whether a constant function between real spaces is smooth and if so, whether the $\psi \circ c \circ \phi^{-1}$ is a consant function defined on a subset of a real space. – Radek Suchánek Jun 22 '17 at 21:02
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Hint: You just need to write it down clearly. $c$ is smooth requires you to show that $\phi \circ c \circ \psi^-$ is smooth. Now as $c$ is constant,for all arguments, you are left with $\phi(y)$ in the end. Now this is just a constant. Use definition of smoothness now

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Every constant map is smooth.

Say $f:M \rightarrow N $ is a constant map. Here $f$ is smooth because the coaordinate representation of $f$ i.e $\theta f\phi^{-1} :R^{m}\rightarrow R^n$ is a constant map as a map between 2 euclidean spaces where $(u,\phi)$ and $(v,\theta)$ are charts around $p$ and $q$ respectively where $f(p)=q$ $\forall$ $p \in$ $M$. Here dimension of $M$ and $N$ are assumed to be $m$ and $n$ respectively. We already know from multi variable calculus that any constant map between two euclidean spaces is always smooth. Hence by definiton of a smooth map between smooth manifolds $f$ is smooth.

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Adittya Chaudhuri
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