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Is the following statement true?

Let $M$ be a compact manifold and $f$ a smooth function from $M$ to itself which is sufficiently close to $id_M$ in $C^1$ topology. Then $f$ is a diffeomorphism.

I know that f is a local diffeomorphism because it is close to the identity in $C^1$ topology, but I could not prove it is injective necessarily.

I need this fact in order to prove for any sufficiently $C^1$ close map $j : M \to T^*M$ to the zero section we have $j(M)=$ image of a one-form. If we know the statement, we can apply it to $\pi oj$ and get the fact that $\pi oj$ is a diffeomorphism. Then $\mu=jo(\pi oj)^{-1}$ is the desired form.

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  • $\begingroup$ this follows from the fact that the set of diffeomorphisms is open in $C^1$ -- i.e. $id$ is an interior point. Rumor has it (see mathoverflow.net/questions/166136/…) that a proof of this can be found in the second chapter of Hirsch's differential topology...(I did not check this source). $\endgroup$
    – Thomas
    Jan 13, 2020 at 18:48

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Here is a sketch of a proof. Fix a Riemannian metric on $M$ and a finite cover of $M$ by compact coordinate charts. By the Lebesgue number lemma, there is $\epsilon>0$ such that every subset of $M$ of diameter $\leq\epsilon$ is contained in one of those coordinate charts. In particular, if $f$ is sufficiently close to the identity in the $C^1$ topology and $f(p)=f(q)$ with $p\neq q$, then there is a single one of the coordinate charts that contains $p$ and $q$ and also contains the entire image under of $f$ of the line segment from $p$ to $q$ (in that coordinate chart). But now looking at $f$ in that coordinate chart, by the mean value theorem there is some point along the line segment where the directional derivative of $f$ along the line segment is perpendicular to the direction of the line segment. In particular, this directional derivative of $f$ is at least a certain distance from the corresponding directional derivative of the identity, which is impossible if $f$ is sufficiently close to the identity in the $C^1$ topology.

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  • $\begingroup$ Thanks for your answer! $\endgroup$
    – ali_ns
    Jan 20, 2020 at 10:57
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This is a reasonably standard compactness argument. Consider any family $\{f_s\}$ of such maps, parametrized by a (small) ball $S=B(0,a)\subset\Bbb R^n$ with $\|f_s-I\|_1<\epsilon<1$ and $f_0=I$. Consider the map $$F\colon M\times S \to M\times S, \quad F(x,t) = (f_s(x),s).$$ Then since $f_s$ is a local diffeomorphism for every $s$, $F$ is an likewise a local diffeomorphism.

Suppose that for every $n\in\Bbb N$, we have a function $f_{s_n}$ with $\|f_n-I\|_0<1/n$ and $f_{s_n}(x_n)=f_{s_n}(y_n)$, $x_n\ne y_n$. By compactness, we pass to subsequences and may assume that $x_n,y_n\to x_0\in M$. But since $F$ is a local diffeomorphism at $(x_0,0)$, there is a neighborhood $U\subset M\times S$ of $(x_0,0)$ on which $F$ is a diffeomorphism. For large enough $n$, we will have $(x_n,s_n),(y_n,s_n)\in U$, contradicting the fact that $F|_U$ is one-to-one.

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  • $\begingroup$ But why must a sequence of counterexamples approaching the identity fit into such a family parametrized by a ball in $\mathbb{R}^n$? $\endgroup$ Jan 13, 2020 at 20:12
  • $\begingroup$ Fair enough, @Eric. So given a particular $f$ close to the identity, I would need to create a (one-parameter) family by moving, as you did in your argument, along geodesics from $p$ to $f(p)$. $\endgroup$ Jan 13, 2020 at 20:16

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