Let $i_{q0} : M\rightarrow M\times N$, $i_{q0}(p) = (p, q0)$ be a mapping between smooth manifolds. I need some hints to show that it is $C^{\infty}$.

I have so far... Let $(U,\phi)$ and $(V,\psi)$ be charts about $p$ and $i_{q0}$, and let $r^{i}$ be the $ith$ coordinate function on Euclidean space. Then we need to show that $\frac{\partial (r^{i}\circ \psi \circ i_{q0} \circ \phi^{-1})}{\partial r^{j}}$ exists and is continuous at $\phi(p)$ and that we can keep taking partial derivatives.

  • $\begingroup$ The map $i_{q_0}$ of your question is simply the inclusion map $(x^1,\dots,x^n)\to (x^1,\dots,x^n,0,\dots,0)$ in (an appropriate choice of) local coordinates where $M$ is a smooth $n$-manifold and $N$ is a smooth $m$-manifold (where $m$ is the number of zeros in $(x^1,\dots,x^n,0,\dots,0)$). Clearly, this (inclusion) map is smooth. $\endgroup$ Feb 16, 2012 at 15:06

2 Answers 2


The smooth structure on $M \times N$ is understood to be the maximal atlas which includes all charts which are products of charts from $M$ and $N$. You do not need to show that the composition of transition maps with your given function are smooth for every choice of charts; you only need to show there is at least one such chart-- this is the point of the axiom which states that overlapping charts are compatible.


I think I got it. Here goes...

Let $(U,\phi)$ be any chart about $p \in M$. Then for any chart $(V,\psi)$ about $q0 \in N$, $(U \times V, \phi \times \psi)$ is a chart about $(p,q0) \in M\times N$. And for $\phi(p) \in R^{m}$, say $\phi(p) = (p^{1},...,p^{m})$, $r^{i}\circ(\phi\times\psi)\circ i_{q0}\circ\phi^{-1}(p^{1},...,p^{m}) = r^{i}\circ((p^{1},...,p^{m}),\psi(q0))$.

Then the partial derivatives are easy to compute after that.

  • $\begingroup$ Or maybe you can first show that the image set is a submanifold of MxN, as, e.g., the zero set of a smooth map, say, define the map as f-x for x in M, and use bump functions to extend it. Then your map is the inclusion map of a submanifold into a manifold, which is smooth when you use the subspace topology for M $\endgroup$
    – gary
    May 19, 2011 at 5:47
  • $\begingroup$ You don't need to compute the partial derivatives. Just observe that for $x\in U$, $(\phi\times\psi\circ i_{q_0}\circ\phi^{-1}(x^1,\cdots,x^m)=(x^1,\cdots,x^m,0,0,\cdots,0)$, which is known, by calculus, to be smooth (I am assuming that $\psi(q_0)=(0,\cdots,0))$. $\endgroup$ Sep 21, 2020 at 14:16

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