Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the disk $D^2$ and its boundary $ \partial D^2 = S^1$. Then $(D^2, S^1 )$ is a good pair. This means there is a boundary map induced

$$ \tilde{H}_2(D^2 / S^1 ) \xrightarrow{\ \partial \ } \tilde{H}_1 (S^1)$$

But we have $D^2 / S^1 \cong S^2$. We also have that $\tilde H_2 (S^2) \cong \mathbb Z$ and $\tilde H _1 (S^1) \cong \mathbb Z $. So the above boundary map corresponds to a map $ \mathbb Z \to \mathbb Z$. I would very much like this map to be multiplication by $2$. Is this true?

In my current course we do not give a proof of what the boundary map $\partial$ is or how to construct it; we just acknowledge that it exists. Is it possible to show that it corresponds to multiplication by $2$ in a simple way?

share|improve this question
2  
That map $\partial$ is part of a long exact sequence; the groups before and after $\partial$ are $\widetilde{H_2}(D^2)$ and $\widetilde{H_1}(D^2)$ -- both vanish. So $\partial$ has to be an isomorphism. –  Thomas Belulovich Nov 5 '12 at 19:51
1  
I'll just add that the map is constructed by starting with the short exact sequence of chain complexes $$ 0\to C_*(S^1) \to C_*(D^2) \to C_*(D^2/S^1) \to 0$$, and then taking the induced long exact sequence on homology. If you have a representing cycle you can then compute the boundary directly. –  Justin Young Nov 5 '12 at 20:03
    
Actually the map $C_*(D^2) \to C_*(D^2/S^1)$ isn't surjective (take a path that starts near $\partial D^2$, runs into this boundary, then exits somewhere else -- you can do this because of the identification, but that map doesn't lift to $D^2$) but $C_*(D^2/S^1)$ is chain homotopic to $C_*(D^2)/C_*(S^1)$. –  Thomas Belulovich Nov 6 '12 at 15:34
    
Yes, I should have said $C_*(D^2, S^1)$ and mentioned excision. –  Justin Young Nov 6 '12 at 17:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.