Proof of the Homotopy Lemma in Topology from the Differentiable Viewpoint

Question

Homotopy Lemma Let $f, g : M \to N$ be smoothly homotopic maps between manifolds of the same dimension, where $M$ is compact and without boundary. If $y \in N$ is a regular value for both $f$ and $g$, then $$\#f^{-1}(y) = \#g^{-1}(y) \ \ \ \ \ (\text{mod}\ 2)$$

Now in his proof there's a line that says the following,

But we recall from Chapter $2$, that a compact $1$-manifold always has an even number of boundary points.

I don't see why this is true, the circle $\mathbb{S}^1$ is a compact $1$-manifold, and it seems to have infinitely many boundary points.

Futhermore, the result that Milnor seems to be quoting, is the classification of $1$-manifolds in the appendix of Topology from the Differentiable Viewpoint. However that result requires, $M$ to be connected, however that is not one of the required conditions on the Homotopy Lemma.

Is this an error in the book? (Also I'm not sure that Chapter $2$ contains the result that Milnor is quoting).

Answer 1

Boundary of a manifold with boundary is different from the topological boundary. This is a crucial point. A circle is a manifold without boundary, hence has $0$ boundary points.

The classification of $1$-manifolds tells you that a connected, compact $1$-dimensional manifold with boundary is either a closed interval or a circle. This then tells you what the components of a general compact $1$-dimensional manifold with boundary must be.

Answer 2

Note: Full credit for this answer must go to Prof. Ted Shifrin (above). This is just a small expansion on his answer for sake of completeness (and so that I don't forget).

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In the appendix of Milnor, it's stated that any smooth, connected $1$-dimensional manifold is diffeomorphic to either the circle $S^1$, or to some interval of real numbers.

One can easily see from this that any smooth, connected, compact, $1$-manifold is diffeomorphic to $[0, 1]$ or $S^1$ (because these are the only compact options that a smooth, connected $1$-manifold can be diffeomorphic to).

Now suppose $M$ was a smooth, compact, but not necessarily connected, $1$-manifold, since every manifold is locally connected, the components of $M$ are open in $M$ and hence also $1$-manifolds, and in any topological space the components of the space are closed, hence the components of $M$ are compact, and they are connected (by the definition of a component), hence they are diffeomorphic either to $[0, 1]$ or $S^1$.

Each component $C \subseteq M$ can easily be seen to have either two boundary points, $\{0, 1\}$ if $C$ is diffeomorphic to $[0, 1]$, or zero boundary points if $C$ is diffeomorphic to $S^1$. Summing all of those boundary points up we get the number of boundary points of $M$, which is thus even.

The fact that adding up all the boundary points of the components of $M$ yields the total number of boundary points in $M$ is intuitive. Take $\left[0, \frac{1}{2}\right] \cup \left[1, 2\right]$, which is a smooth, compact, disconnected $1$-manifold as a prototypical example.