Given an orientation-preserving diffeomorphism $h: \partial D^m \to \partial D^m$, we can glue two copies of the closed unit disk $D^m$ along the boundary by identifying $x \sim h(x)$ to form the quotient space $$\Sigma(h) := (D^m \amalg D^m)/\sim$$ Now we can give this quotient a smooth structure such that the obvious inclusions $D^m \hookrightarrow \Sigma(h)$ are smooth embeddings and in fact it turns out that for any two smooth structures, there exists a diffeomorphism between them. So $\Sigma(h)$ is a unique manifold up to diffeomorphism.
So far, so good. Now, in Kosinski's 'Differential Manifolds', there is the following Lemma:
Lemma: $\Sigma(h)$ is diffeomorphic to $S^m$ if and only if $h$ extends over $D^m$. Moreover, $\Sigma(gh) = \Sigma(h)\# \Sigma(g)$.
Here $M\# N$ denotes the connected sum of two manifolds as usual.
The proof of this is left as an exercise for the reader, but I'm unsure, how one might construct an extension of $h$ to $D^m$, given that $\Sigma(h)$ is diffeomorphic to $S^m$?
I know that in this case $h(\partial D^m)$ necessarily separates $\Sigma(h) = S^m$ into two components, and since $h(\partial D^m)$ is an embedded compact $(m-1)$-manifold (which is smooth), I can also prove that $h(\partial D^m)$ is the boundary of both connected components of its complement.
But at this point I get lost. Is it clear that these two components are diffeomorphic to disks? Where might I find a proof of this?
I'm fine with the other parts of this Lemma, but I just don't see how to extend $h$ given $\Sigma(h) = S^m$. If you could help me out, this would be very much appreciated. Thank you for your help!