# Proving that the boundary of all compact $1-$ manifolds have an even number of points.

It is well-known that if $$M$$ is a connected $$1-$$ manifold, then it is diffeomorphic to either $$[0,1]$$, $$[0,1)$$, $$(0,1)$$, or $$\mathcal{S}^{1}$$.

It is often stated as a trivial corollary that if, in addition, $$M$$ is compact, then the boundary $$\partial{M}$$ has an even number of points. My question is how this follows from the above statement.

It is clear that if $$M$$ is compact, then that limits the choices which $$M$$ can be diffeomorphic to to either $$[0,1]$$ or $$\mathcal{S}^{1}$$. So let's suppose that it is the case where $$M$$ is diffeomorphic to $$[0,1]$$ through the map $$\phi:[0,1] \rightarrow M$$.

Now, it's clear that $$(\phi, [0,1])$$ is not sufficient to be a chart for every point in $$M$$, because $$[0,1]$$ is not open in either $$\mathbb{R}$$ nor the halfspace $$H^{1}$$. Regardless, $$M$$ is a $$1-$$manifold by way of some charts by assumption, and hence we can talk about $$int(M)$$ and $$\partial{M}$$. We know from standard results that $$int(M)$$ is open in $$M$$ and $$\partial{M}$$ is closed in $$M$$, and that both sets are disjoint from each other, so we can say that

$$\phi: [0,1] \rightarrow int(M) \cup \partial{M}$$

is a diffeomorphism. But at this point I am a bit stuck. I feel like we should be able to prove that $$\partial{M}=\phi(\{0,1\})$$, and hence $$\partial{M}$$ consists of $$2$$ points, an even number, but I can't seem to make any headway here.

Can anybody point me in the right direction? Thanks!

• Step a little back from the problem at hand, and think about "If $f \colon M \to N$ is a diffeomorphism between manifolds with boundary, then $f$ maps the interior of $M$ to the interior of $N$, and $\partial M$ to $\partial N$." Would that solve your problem? Can you see how to go about proving this? Jun 6, 2020 at 19:28
• $0$ is a boundary point of $[0,1]$ with boundary chart $[0,1)$. SInce $\phi$ diffeomorphism, $\phi([0,1))$ is a boundary chart for $\phi(0)$. By invariance of domain $\phi(0)$ is in $\partial M$. Likewise for $1$. Jun 6, 2020 at 19:33

I think I'm probably repeating what is said in the comments. We know that a compact connected manifold (with boundary) of dimension $$1$$ is diffeomorphic to either $$[0,1]$$ or $$S^1$$. Note also that diffeomorphisms of manifolds with boundary preserve the boundary, i.e. if $$f:M\to N$$ is a diffeomorphism of manifolds with boundary $$f(\partial M)=\partial N$$. So, if $$\#\partial M$$ is finite, then so is $$\#\partial N$$ and $$\#\partial M=\#\partial N$$ when $$M\cong N$$.
It is not too hard to see that $$\partial([0,1])=\{0\}\cup \{1\}$$, and hence that $$\#\partial [0,1]=2$$. On the other hand, $$\partial S^1=\varnothing$$, so that $$\#\partial S^1=0$$. So, we have proven the result in dimension $$1$$.
The classification of compact connected $$1-$$dimensional manifolds with boundary extends to the disconnected case, so that a compact $$1-$$dimensional manifold is diffeomorphic to $$M(k,\ell)=\bigg(\coprod_{i=1}^k [0,1]\bigg)\sqcup\bigg(\coprod_{j=1}^\ell S^1\bigg)$$ for some $$k,\ell\ge 0$$. Then, we see that $$\#\partial M(k,\ell)=2k$$. So, the result follows.
• Thank you! I think I understand the case when $M$ is connected now. In order to achieve the disconnected result, does it follow immediately from the following facts: 1. A compact (disconnected) manifold has a finite number of connected components, each of which is an open. 2. Therefore, by breaking up $M$ into it's finite components, we simply add up all of the boundary points for each of components in order to get the boundary for $M$ itself. Or is it more complicated than that?