I have a question about an proof in Milnor's TOPOLOGY FROM THE DIFFERENTIABLE VIEWPOINT (see pages 65-66): The aim is to show the classifying theorem that any smooth, connected $1$-dimensional manifold is difeomorphic either to the circle $S^1$ or to some interval of real numbers.

In order to show it the author uses following lemma:

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Here the proof with red tagged argument which isn't clear to me:

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We take the graph $\Gamma \subset I \times J$ consisting of all $(s,t)$ with $f(s)= g(t)$ ($f, g$ parametrisations; for the notation: see above)

My questions are following:

  1. Why $\Gamma$ is closed in $I \times J$? (my considerations: I guess that because for small enough open $U \subset M$ the diagonal of $U \times U$ is closed (since $M$ Hausdorff) and $\Gamma$ is just it's preimage. Is the argument ok?)

  2. Why the lines of $\Gamma$ cannot end in the interior $\mathring{I} \times \mathring{J}$? Why does the fact that $g^{-1} \circ f$ is a local isomorphism exclude it?


$\Gamma$ is closed since $f\times g$ is continuous, $\Gamma=(f\times g)^{-1}(D)$ where $D=\{(x,x)\}$ is the diagonal.

Suppose that a segment $c$ end to the interior, there exists a family $(u_n,v_n)$ with $f(u_n)=f(v_n)$ and $lim_n(u_n,v_n)=(u,v)$ is the end of $c$, since $f,g$ are continuous, $(f\times g)(u,v)=lim_n(f(u_n),g(v_n))$ implies that $f(u)=f(v)$. Thus we can assume that the segment is closed. We have $g(^{-1}\circ f)(u)=v$ and $u$ is in the interior of $I$, Write $u_t=u+t$ where $u+t$ is in $I$, write $v_t=(g^{-1}\circ f)(u_t)$, we have $(u_t,v_t)$ extends $c$.

  • $\begingroup$ Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}\circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) \in im(g)$? Obvioulsly it suffice to show that $f(u) \in \overset{\circ}{im(g)}$. Does for a parametrisation always hold following statement: $t \in \mathring{J} \Leftrightarrow g(t) \in \mathring{im(g)}$ $\endgroup$ – KarlPeter Jan 18 at 20:25

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