# C1-close maps to identity

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.

• 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). Jan 13, 2020 at 18:48

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.
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.
• But why must a sequence of counterexamples approaching the identity fit into such a family parametrized by a ball in $\mathbb{R}^n$? Jan 13, 2020 at 20:12
• 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 $f(p)$. Jan 13, 2020 at 20:16