Attaching Two Discs Along the Boundary but what about differentiability?

Question

I have seen this post. If we provide a diffeomorphism $h:\partial \mathbb{B}^2\to\partial\mathbb{B}^2$ and glue along the boundary with it, we get a space that is diffeomorphic with $S^2$? The construction in that post is not differentiable at $S^1\subset S^2.$

Answer 1

Proposition. Every degree 1 diffeomorphism of $S^1$ is smoothly isotopic to the identity.

Proof. First, some generalities. Let $M$ be a connected manifold and $f: M\to M$ a continuous map. This map induces a homomorphism $f_*: \pi_1(M,m)\to \pi_1(M, f(m))$. The identification $\pi_1(M,m)\cong \pi_1(M,m')$ for different base-points is not canonical, it is unique up to conjugation. Thus, in general, $f$ induces a homomorphism $\phi: \pi_1(M)\to \pi_1(M)$ well-defined up to an inner automorphism of $\pi_1(M)$. If $G=\pi_1(M)$ is abelian (e.g. $M=S^1$), which I will assume from now on, then every inner automorphism of $G$ is trivial, hence, we have a well-defined homomorphism $f_*: G\to G$. Let $p:\tilde{M}\to M$ be the universal covering; the group of covering transformations of $p$ is isomorphic to $G$. The map $f$ lifts to a map $\tilde{f}: \tilde{M}\to \tilde{M}$ which induces the homomorphism $\phi$: $$ \tilde{f}\circ g= \phi(g)\circ \tilde{f}, \forall g\in G. $$ Conversely, if $F: \tilde{M}\to \tilde{M}$ is a map satisfying the above identity, then $F$ projects to a map $f: M\to M$ and the projection $f$ induces the homomorphism $\phi: \pi_1(M)\to \pi_1(M)$. If $f$ is homotopic to the identity map, then $\phi=id_G$; accordingly, $$ \tilde{f}\circ g= g\circ \tilde{f}, \forall g\in G. $$

Now, let us specialize to the case of $M=S^1$, $p: {\mathbb R}\to S^1$, $p(x)= e^{2\pi ix}$, where I am identifying $S^1$ with the unit circle in the complex plane. The group $G$ of covering transformations of $p$ is generated by the translation $g_1: x\mapsto x+1$. Therefore, $f: S^1\to S^1$ has degree 1 if and only if it is homotopic to the identity, if and only if its lift $\tilde{f}$ satisfies $$ \tilde{f}(x+ 1)= \tilde{f}(x) +1, \forall x\in {\mathbb R}. $$ For such a lift we define the straight-line homotopy to the identity: $$ \tilde{F}(x,t)= (1-t) \tilde{f}(x) + t x. $$ By a direct calculation one verifies the identity $$ \tilde{F}(x+ 1,t)= \tilde{F}(x,t) +1, \forall x\in {\mathbb R}, t\in [0,1]. $$ Therefore, $\tilde{F}(x,t)$ projects to a homotopy $F(z,t)$ from $f(z)$ to $id_{S^1}$.

Assume now that $f$ is also differentiable, thus, $\tilde{f}$ is also differentiable and we obtain $$ \frac{\partial }{\partial x} \tilde{F}(x,t)= (1-t)\tilde{f}'(x) + t. $$ If $\tilde f'(x)>0$ for all $x$, then $\frac{\partial }{\partial x} \tilde{F}(x,t)>0$ for all $x$ and all $t\in [0,1]$ as well. Suppose furthermore, that $f$ is a diffeomorphism, so is $\tilde{f}$; hence, $\tilde{f}'(x)\ne 0$ for all $x$. If $\tilde{f}'(x)< 0$ for all $x$, then $\tilde{f}$ is decreasing and, hence, cannot satisfy the identity $$ \tilde{f}(x+ 1)= \tilde{f}(x) +1. $$ Thus, $f'(x)>0$ for all $x$ and, hence, $\frac{\partial }{\partial x} \tilde{F}(x,t)>0$ for all $x$ and all $t\in [0,1]$. Thus, for each $t$, the map $\tilde{F}(x,t)$ is a local diffeomorphism of the real line. (One verifies its subjectivity using the same identity $$ \tilde{F}(x+ 1,t)= \tilde{F}(x,t) +1, \forall x\in {\mathbb R}, t\in [0,1] $$ and, hence, for each $t$, the map $\tilde{F}(x,t)$ is a diffeomorphism, but I will not need this.) Thus, for each $t$ fixed, the map $\tilde{F}(x,t)$ descends to a local diffeomorphism $F(z,t)$, $S^1\to S^1$. Since $S^1$ is compact, this local diffeomorphism is a covering map. This covering map has to be a degree 1 covering map since it is homotopic to $f$. Thus, we obtain the required isotopy of $f$ to the identity through diffeomorphsms of $S^1$. qed