# Restriction of diffeomorphisms homotopic to identity to the boundary

Let $$M$$ be a smooth manifold with boundary $$\partial M$$. Let $$Diff_0(M)$$ be the group of all diffeomorphisms homotopic to identity. According to this article (Page 6, section " Beyond mapping class group"), the restriction of a diffeomorphism to the boundary gives a well defined surjactive homomorphism $$\phi: Diff_0(M)\rightarrow Diff_0(\partial M).$$

I could not find a reference for the above two results. So I have two questions.

1) Why is $$\phi$$ well defined, i.e., why the restriction of a difeomorphism homotopic to identity in $$M$$ is homotopic to identity in $$\partial M$$.

2) Why every diffeomorphism homotopic to identity in $$\partial M$$ is a restriction of a diffeomorphism of $$M$$ which is homotopic to identity.

As I am not quite familiar with diffeomorphism groups, any suggestion/reference/comment will be extremely helpful. Also I would request you to improve the tags if possible.

• Path component of the identity means isotopy. To get 1, just restrict the isotopy to the boundary, it remains isotopy. (The difference with homotopy is that boundary maps to boundary.) To get 2, use the existence of a collar neighborhood of the boundary which is the product $\partial M\times [0,1]$. May 9 '20 at 4:23
As discussed in comments, $$Diff_0$$ stands for the subgroup of the diffeomorphism group, consisting of diffeomorphisms isotopic (rather than homotopic) to the identity. With this in mind, the fact that the restriction map $$\phi: Diff(M)\to Diff(\partial M)$$ sends $$Diff_0(M)$$ to $$Diff_0(\partial M)$$ is clear. Let's prove surjectivity. First of all, $$\partial M$$ admits a "collar" $$C$$ in $$M$$, a closed neighborhood of $$\partial M$$ in $$M$$, $$C$$ is diffeomorphic to $$\partial M\times [0,1]$$. Now, given $$h\in Diff_0(\partial M)$$, let $$H(x, t), t\in [0,1]$$, denote the isotopy of $$h= H(\cdot, 0)$$ to $$id_{\partial M}= H(\cdot, 1)$$. I leave it to you to prove that $$H$$ can be chosen so that $$H(x,t)=x$$ for all $$t\in [1/4, 1]$$. Then, using the diffeomorphism $$C\cong \partial M\times [0,1]$$, extend $$h$$ first to $$C$$ and then, by identity, to the rest of $$M$$. Call the extension $$\hat{h}$$. Clearly, $$\phi(\hat{h})=h$$. It remains to prove that $$\hat{h}\in Diff_0(M)$$. To prove this, play the same game as before: Given an isotopy $$H(x,t)$$ from $$h$$ to $$id_{\partial M}$$, extend it to $$C\cong \partial M\times [0,1]$$ by $$(x,t,s)\mapsto H(x, t+s),$$ and then by identity to the rest of $$M$$. This will be an isotopy $$\hat{H}$$ from $$\hat{h}$$ to $$id_M$$.
Here is an example to ponder: Let $$M$$ be the annulus $$S^1\times [0,1]$$. Consider the diffeomorphism $$f(s,t)=(s, 1-t)$$; $$f: M\to M$$ is homotopic to the identity, but its restriction to $$\partial M$$ is not.