0
$\begingroup$

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:

enter image description here

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

enter image description here

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?

$\endgroup$
1
$\begingroup$

$\Gamma$ 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 $c$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.