4
$\begingroup$

I'm self-studying Munkres' "Elements of Algebraic Topology". And I'm thinking Section36 Exercise #2 and looking for hints or solution. Here it goes,

Section36 Exercise#2

Let $$Y=\{0\}\times [-1,1]\bigcup\{(x,\sin{\frac{1}{x}})\colon x\in(0,1]\}$$ $$Z=\textit{Union of }\space Y\textit{ and an arc that intersects }Y\textit{ only in the points }(0,-1)\textit{ and }(1,\sin{1})$$ Then, verify following statements

(a) If $h\colon Y\rightarrow S^n$ is an imbedding, then $S^n-h(Y)$ is acyclic.(i.e, every reduced singular homology group is trivial.)

(b) If $h\colon Z\rightarrow S^n$ is an imbedding, then reduced singular homology group $\widetilde{H}_i(S^n-h(Z))$ is infinite cyclic if $i=n-2$, and vanishes otherwise.

(c) If $h\colon Z\rightarrow S^2$ is an imbedding, then $S^n-h(Z)$ has precisely two components, of which $h(Z)$ is the common boundary

How can I approach these three problems?

-It seems above exercises are all somehow related to Jordan curve theorem and its related theorem.(anyhow, Section36 is about Jordan curve theorem). However, $Y$ is not homeomorphic to $[0,1]$ and $Z$ is not homeomorphic to $S^1$. So we cannot directly apply those theorems to here.

Thanks for any help in advance.

$\endgroup$
  • $\begingroup$ Is it $\sin( 1/x)$ instead ? $\endgroup$ – user99914 Oct 7 '16 at 9:48
  • $\begingroup$ @John Ma oops..you are correct. editted. Thanks! $\endgroup$ – KGEO Oct 7 '16 at 12:58
  • 3
    $\begingroup$ Do you know about Chech cohomology and Alexander duality? Once you learn these, the solution is quite simple. Incidentally, part (c) is incorrect, you need $n=2$. $\endgroup$ – Moishe Kohan Oct 10 '16 at 13:59
  • $\begingroup$ @Moishe Cohen Thanks for the correction. I've just editted. And no, I didn't learn Chech cohomology and Alexander duality yet. Maybe I should learn more materials, then. But, still If I can find elementary solution that would be good. $\endgroup$ – KGEO Oct 11 '16 at 7:29
  • 1
    $\begingroup$ My guess is that Munkres wants you to to imitate his proof of Jordan separation theorem. This seems quite painful to me. $\endgroup$ – Moishe Kohan Oct 11 '16 at 11:08

Your Answer

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

Browse other questions tagged or ask your own question.