# Differentiable injective function betweem manifolds

Let $f:M\rightarrow N$ an injective differentiable functions between the manifolds M and N. Prove that $\dim M\leq \dim N$.

Can anyone give me a hint to solve this theorem? I tried to relate the injectivity of $f$ with the injectivity of its differential but that is not gonna work.

Thanks!

• I really dont understand why my question is downvote. – YTS Aug 30 '17 at 17:42

Let $m:=\dim(M)$ and $n:=\dim(N)$. Let us prove the contrapositive and assume that $m>n$.

• Step 1. Reduction to $M=\mathbb{R}^m$ and $N=\mathbb{R}^n$.

Let $x\in M$, let $(U,\phi)$ be a chart of $M$ centered at $x$ and let $(V,\psi)$ be a chart of $N$ centered at $f(x)$, then let define the following map: $$g:=\psi\circ f\circ\phi^{-1}\colon\phi(f^{-1}(V)\cap U)\rightarrow\mathbb{R}^n.$$ The domain of $g$ is open and non-empty in $\mathbb{R}^m$, so that it contains a euclidean ball, let's called it $B$. Now, recall that $B$ is diffeomorphic to $\mathbb{R}^m$, let $\varphi\colon\mathbb{R}^m\rightarrow B$ be one of them. Notice that: $$h:=g\circ\varphi\colon\mathbb{R}^m\rightarrow\mathbb{R}^n$$ is a smooth map which is injective if and only if $f$ is.

• Step 2. Conclusion.

Since $n\leqslant m-1$, let $i\colon\mathbb{R}^n\rightarrow\mathbb{R}^{m-1}$ be the inclusion, namely the map defined by: $$i(x_1,\ldots,x_n):=(x_1,\ldots,x_n,0,\ldots,0).$$ Then, notice that: $$(i\circ h)_{\vert\mathbb{S}^{m-1}}\colon\mathbb{S}^{m-1}\rightarrow\mathbb{R}^{m-1}$$ is a smooth map which cannot be injective as it admits an antipodal point by Borsuk-Ulam theorem. Therefore, since $i$ is injective, $h$ is not.

Given a smooth injective map from $\mathbb{R}^m$ to $\mathbb{R}^n$, can one build a smooth surjective map from $\mathbb{R}^n$ to $\mathbb{R}^m$ ? If will assume that it is true.

Let $h\colon\mathbb{R}^n\rightarrow\mathbb{R}^m$ be a smooth map, since $\mathbb{R}^n$ is Lebesgue-negligible in $\mathbb{R}^m$, using mean value inequality, $h(\mathbb{R}^n)$ is also Lebesgue-negligible in $\mathbb{R}^n$. Whence, $h$ is not surjective.

• What kind of Borsuk-Ulam do you use. pt.wikipedia.org/wiki/Teorema_de_Borsuk-Ulam – YTS Aug 30 '17 at 17:39
• I think that I understand what you mean, but is it not possible to solve this really using the smoothness of $f$? You are merely using its properties as a continuous functions. – YTS Aug 30 '17 at 17:40
• Here is another possible approach. – C. Falcon Aug 30 '17 at 18:01
• Can you explain why you can just consider $R^{m}$ and $R^{n}$, if the charts doesnt have to send to all the space, just to an open set. And please define the inclusion in spaces of possibly different dimensions. – alexp9 Aug 31 '17 at 12:03
• @Rhcpy99 Here you have some more details, which are in my opinion just tedious and give no more insight. – C. Falcon Aug 31 '17 at 17:51