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This question already has an answer here:

Let $M$ and $N$ be two smooth manifolds with dimensions m and n, respectively, with $m>n$. Let $f:M\to N$ a smooth map ($C^\infty$). Show that this map can't be injective.

My try: I was wondering how could I reduce this problem to the tangent space, using the differentiability condition, to work with vector spaces, where I know how to proof this result. But I was unable to do it, so any help will be useful.

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marked as duplicate by C. Falcon, user91500, user354271, José Carlos Santos, Frpzzd Aug 31 '17 at 23:51

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Among all points of $M$, choose a point $x$ where the rank of the differential $d_x f$ is biggest. It then follows that in some small open neighborhood of $x$, the function $f$ has constant rank. Then by the constant rank theorem it'll be obvious that $f$ cannot be injective there.

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  • $\begingroup$ +1 as it is much easier than my approach! I will emphasize on the importance to take a point $x$ where $\textrm{rank}(\mathrm{d}_xf)$ is maximal to conclude that $f$ has constant rank near $x$. This follows from the fact that locally the rank can only grow as minors of a matrix are continuous. $\endgroup$ – C. Falcon Aug 30 '17 at 23:47
  • $\begingroup$ Can you explain why the constant rank theorem gives the desired result? $\endgroup$ – alexp9 Aug 31 '17 at 18:41
  • $\begingroup$ @Rhcpy99 Let $r$ be the rank of $\mathrm{d}_xf$, one has $r\leqslant\dim(N)<\dim(M)$ and according to the constant rank theorem there exists diffeomorphisms $\varphi,\psi$ such that: $$(\varphi\circ f\circ\psi)(x_1,\ldots,x_m)=(x_1,\ldots,x_r,0,\ldots,0)$$ in a neighborhood of $x$. Therefore, $\varphi\circ f\circ\psi$ is not injective and so is $f$. $\endgroup$ – C. Falcon Aug 31 '17 at 19:53
  • $\begingroup$ We actually have $m > n$, so that the function is more like $(x_1, \ldots, x_m) \to (x_1, \ldots, x_n)$. It's not injective because it forgets some coordinates. $\endgroup$ – Pedro Sep 1 '17 at 1:20
  • $\begingroup$ @Pedro This will happen if and only if $f$ has full rank, but perhaps there is no $x$ such that $\textrm{rank}(\mathrm{d}_xf)=n$. Anyway, $(x_1,\ldots,x_m)\mapsto (x_1,\ldots,x_r,0,\ldots,0)$ also forgets some coordinates as $r\leqslant n<m$. $\endgroup$ – C. Falcon Sep 1 '17 at 10:53

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