# A Lemma from Milnor's Topology from the Differentiable Viewpoint

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

## 1 Answer

$$\Gamma$$ 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 $$c$$.

• 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: $t \in \mathring{J} \Leftrightarrow g(t) \in \mathring{im(g)}$ – KarlPeter Jan 18 at 20:25