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I was trying to compute the homology of a torus with the long exact sequence for relative homology formed quotient, inclusion, and boundary $ \dots\to\tilde{H}_n(A)\overset{i_*}{\to} \tilde{H}_n(X)\overset{j*}{\to}\tilde{H}_n(X/A)\overset{\partial}{\to}\tilde{H}_{n-1}\to\dots $
where $A\subseteq{}B$, $i_*$ is the homomorphism induced by inclusion, and $j_*$ is the homomorphism induced by quotienting.

I figured a torus should have the same homology groups as a hollow cylinder modulo its ends (kind of like this http://inperc.com/wiki/images/8/8f/Torus_via_gluing.jpg)

So by letting $X$ = the cylinder and $A$ = its ends, the homology of a torus should be $\tilde{H}_n(X/A)$.

The reduced homology of a cylinder would be $\tilde{H}_0 = 0$ $\tilde{H}_1 = \mathbb{Z}$ $\tilde{H}_{n>1} = 0$

The reduced homology of A would be $\tilde{H}_0 = \mathbb{Z}$ $\tilde{H}_1 = \mathbb{Z}^2$ $\tilde{H}_{n>1}=0$

The reduced homology of a torus should be $\tilde{H}_0 = 0$ $\tilde{H}_1 = \mathbb{Z}^2$ $\tilde{H}_2=\mathbb{Z}$ $\tilde{H}_{n>2}=0$

Then $ 0\overset{j_*}{\to}\tilde{H}_2(X/A)\overset{\partial}{\to}\tilde{H}_1(A)\overset{i_*}{\to}\tilde{H}_1(X)\overset{j_*}{\to}\tilde{H}_1(A/X)\overset{\partial}{\to}\tilde{H}_0(A)\overset{i_*}{\to}\tilde{H}_0(X) $
which by substitution is
$ 0\overset{j_*}{\to}\mathbb{Z}\overset{\partial}{\to}\mathbb{Z}^2\overset{i_*}{\to}\mathbb{Z}\overset{j_*}{\to}\mathbb{Z}^2\overset{\partial}{\to}\mathbb{Z}\overset{i_*}{\to}0 $
should be a short exact sequence, but I keep getting the image of the second $j_*$ to be $0$, which would mean there's an isomorphism from $\mathbb{Z}^2\to\mathbb{Z}$.

Can you help me find the mistake? Thanks!

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I didn't follow your whole proof, but one problem is that your $X/A$ is not homeomorphic to the torus. If you glue the circle at one end of the cylinder $X$ to the circle at the other end in the correct orientation, you do get the torus. But $X/A$ collapses the whole subspace $A$ to a single point, which is different: it's a torus with one meridian pinched together. My guess is that your $X/A$ has the same first homology group as a circle, which might explain your results.

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That's what I was thinking too. I guess the pinched meridian takes away a homology class and I didn't realize. –  brian Aug 20 '12 at 23:49
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I think the $X/A$ should come out to be the same as a sphere with north and south pole identified, which is homotopy equivalent to $S^2 \vee S^1$. –  Thomas Belulovich Aug 21 '12 at 0:06
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