15
$\begingroup$

It is one of the problems in Hatcher's book.

I need to find the homology group of $H_n (X,A)$ when $A$ is a finite set of points and $X$ is $S^2$ or $T^2$.

I figured out that for $n>1$, I could use the long exact sequence and make $H_n (X,A)$ isomorphic to $H_n (X)$.

However, I am stuck with $H_1 (X,A)$ and $H_0 (X,A)$.

Can anyone give me an idea how I can find these?

$\endgroup$
4
  • 4
    $\begingroup$ For $H_0 (X,A)$ (in both cases, $S^2$ and $T^2$) use that $H_n(X,A) \cong \tilde{H_n}(X/A)$ (Hatcher, page 124). You know that $X/A$ is path-connected so you get $H_n (X/A) = \mathbb Z$ and hence $H_n(X,A) = 0$. $\endgroup$ Apr 9, 2012 at 17:12
  • $\begingroup$ thanks! that gives me half of the answer! :-) $\endgroup$
    – Emily
    Apr 10, 2012 at 15:27
  • $\begingroup$ Welcome : ) Glad I could help. $\endgroup$ Apr 10, 2012 at 15:54
  • $\begingroup$ @RudytheReindeer Can I ask how we prove that the pair (X,A) is a good pair? It seems weird because A is not connected ... $\endgroup$
    – sifsa
    May 6, 2016 at 15:27

2 Answers 2

13
$\begingroup$

Suppose $A=\{x_1,\dots,x_k\}$. Then $H_0(A)=\mathbb{Z}^k$ and $H_i(A)=0$ for $i>0$.

Whether $X=S^2$ or $T^2$ we have $H_0(X)\cong\mathbb{Z}$, and like Matt N said in his comment in either case $H_0(X,A)\cong\tilde{H}_0(X/A)=0$.

If $X=S^2$ then $H_1(X)=0$ so the l.e.s. has a portion like $$0\rightarrow H_1(X,A)\rightarrow \mathbb{Z}^k\rightarrow \mathbb{Z}\rightarrow 0 $$ and so $H_1(S^2,A)\cong\mathbb{Z}^{k-1}$.

If $X=T^2$ then $H_1(X)=\mathbb{Z}^2$, so we have $$ 0\rightarrow \mathbb{Z}^2\rightarrow H_1(X,A)\stackrel{\partial}{\rightarrow} \mathbb{Z}^k\rightarrow \mathbb{Z}\rightarrow 0$$ Then $\ker\partial\cong\mathbb{Z}^2$ and its image is $\cong\mathbb{Z}^{k-1}$. I believe this is enough to conclude $H_1(T^2,A)\cong \mathbb{Z}^{k+1}$

$\endgroup$
9
  • $\begingroup$ thank you for a generous explanation! :-) But I am still confused about that last sentence. I guess you have used the splitness of the exact sequence. However, how do we know that the sequence splits? If we have a short exact sequence then, $\mathbb Z$ being free abelian would be enough, but can we extend this idea to the long exact sequence as above? $\endgroup$
    – Emily
    Apr 10, 2012 at 15:55
  • 2
    $\begingroup$ My reasoning is like this: $H_1(X,A)$ is a finitely generated abelian group, so which one is it? We know it has no torsion, since the image and kernel of $\partial$ are both free, so it is completely determined by the rank. But then the rank has to be $(k-1)+2$ $\endgroup$
    – William
    Apr 10, 2012 at 17:15
  • $\begingroup$ (hopefully that reasoning is valid...) $\endgroup$
    – William
    Apr 10, 2012 at 17:25
  • $\begingroup$ That explains to me! But can't we say that im(round) is enough for $H_1 (X,A)$ being free? I am asking this to make sure I am understanding right. $\endgroup$
    – Emily
    Apr 10, 2012 at 17:45
  • $\begingroup$ Well, we know that the image of $\partial$ is in a free abelian group, so if $H_1(X,A)$ has torsion it will be in the kernel of this homomorphism. But exactness tells us that $\ker\partial\cong\mathbb{Z}^2$, which has no torsion. $\endgroup$
    – William
    Apr 10, 2012 at 22:35
5
$\begingroup$

This uses Proposition 2.22 in Hatcher(and you must prove $(X,A)$ is a good pair).

Without saying too much, although I guess "you" did, I solved this by finding a homotopy equivalence between the space $S^2/A$ or $T^2/A$ and a CW complex. In the former case, you can see the homotopy between $S^2/A$ and a CW complex given by $k+1$ 0-cells(where $A$ is a collection of $k$ points), $2k$ 1-cells, and 2 2-cells. Graphically we arrange the first $k$ 0-cells into a lovely regular $k$-gon, with an outlier $x$ in the back. We use the first $k$ 1-cells to add edges to our $k$-gon, and the second $k$ connecting the vertices of the $k$-gon to the outlier. Then the two cells are each attached with their boundaries glued to the $k$-gon. By contracting the 1-cells attached to $x$, we see this is homotopic to $S^2/A$, but this has a CW complex structure, so it is easier to compute.

$\endgroup$
2
  • $\begingroup$ This is an interesting point of view! It became geometrically clear, thank you! :-) $\endgroup$
    – Emily
    Apr 10, 2012 at 16:09
  • $\begingroup$ Glad you enjoyed it! $\endgroup$ Apr 10, 2012 at 16:39

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.