i have really big problems in computing the homology groups and even don't really know how to write the exact sequence for different spaces.
i tried to take the torus with n points removed,homotopic to a circle with n circles attached around and use my intuition to get the answer which i'm not sure if is true…
can someone please give a detailed answer? i really want to learn this subject.
[Math] how to compute the homology groups of a torus with n points removed
algebraic-topology
Related Solutions
Although you've got the answer by yourself, I would like to write an answer solving the problem with cellular homology, so that someone who asks the same question can find an answer here. I solved this problem a few month ago in an Algebraic Topology course as an exercise.
Proof:
Let $X = S^1\times S^1/ \sim$ be the space with the identifications:
$$(e^{2\pi i/m}z,x_0)\sim (z,x_0)$$
$$(x_0,e^{2\pi i/n}z)\sim (x_0,z)$$
Like @Berci said, you should imagine this space as a grid of $m$ and $n$ lines, i.e. there are $m$ vertical and $n$ horizontal repititions:
(OK. The picture is not the nicest one, but it's enough to induce an imagination.)
$X$ consists of one 0-cell ($x_0$ is $e_1^0$), two 1-cells ($a$ is $e_1^1$, $b$ is $e_2^1$) and one 2-cell (we all it $e_1^2$).
The attaching map identifies $x\in \partial D_1^2$ with $a^nb^ma^{-n}b^{-m}$.
This implies the cellular chain complex
$$0\to \mathbb{Z}[e_1^2]\overset{\partial_2=0}{\longrightarrow} \mathbb{Z}[a] \oplus\mathbb{Z}[b]\overset{\partial_1=0}{\longrightarrow} \mathbb{Z}[x_0]\to 0.$$
This implies
$$H_p(X) = \begin{cases} \mathbb{Z}\mbox{ for } p=0,2 \\ \mathbb{Z}^2\mbox{ for } p=1 \\ 0\mbox{ for } p>2 \end{cases}.$$
Otherwise you can just see, that the space $X$ is still a Torus (cf. remark above). So it is not surprising, that we've got the homology group of the Torus.
I think people should ignore all this nonsense about counting holes and just look at what actually happens in more examples.
In particular, your intuition about top homology having something to do with enclosing volumes is not quite correct. I interpret this to mean that you have in mind a manifold which is the boundary of another manifold (the same way that the sphere $S^n$ is the boundary of the disk $D^{n+1}$), and it's not true that a manifold has to be a boundary in order to have nonvanishing top homology. The simplest closed counterexample is $4$-dimensional: the complex projective plane $\mathbb{CP}^2$ is a $4$-manifold with $H_4 = \mathbb{Z}$, but it's known not to be the boundary of a $5$-manifold.
Whether top homology vanishes or not instead has to do with orientability.
It's possible to directly visualize the case of $\mathbb{RP}^2$, so let's do it. In this case $\pi_1 \cong H_1 \cong \mathbb{Z}_2$, so the goal is to visualize why there's some loop which isn't trivial but such that twice that loop is trivial. Visualize $\mathbb{RP}^2$ as a disk $D^2$, but where antipodal points on the boundary have been identified. We'll consider loops starting and ending at the origin.
I claim that a representative of a generator of $\pi_1 \cong H_1$ is given by the loop which starts at the origin, goes up to the boundary, gets identified with the opposite point, and goes up back to the origin. Try nudging this loop around for a bit so you believe that it's not nullhomotopic: the point is that you can't nudge it away from the boundary because the two (antipodal, hence identified) points it's intersecting the boundary at can never annihilate.
Now we want to visualize why twice this loop is nullhomotopic. It will be convenient to nudge the loop so that it hits the boundary at four points, which come in two antipodal pairs $A, A', B, B'$, so that the loop hits them in that order before returning to the origin. At this point it would be helpful to draw a diagram if you haven't already; the disk, and the two loops in it, should look a bit like a tennis ball in profile. Now, nudge the loop so that $A, B'$ get closer together, and hence, since they're constrained to be antipodes, $A', B$ also get closer together. Eventually you'll have nudged them enough that you'll see that you can finally pull the curve away from the boundary: as you do so, $A', B$ annihilate each other, and then $A, B'$ annihilate each other.
A similar visualization works for the Klein bottle, visualized as a square with its sides identified appropriately.
Best Answer
After removing $1$ point, and seeing the torus as a $CW$ complex (given by making the usual identifications $I^2/\sim$ of a square) then before gluing, we can enlarge the hole to the boundary of the square. After making identifications, this gives the wedge of two circles.
One can continue in this way to see that the torus with $n$ points removed gives the wedge of $n+1$ circles. In doing this, we can compute the (reduced) homology with Mayer-Vietoris (which is done in Hatcher, if I remember correctly.)
The easiest way to do this is probably induction on the number of circles, and noting that for the wedge of two circles, you can find neighborhoods so that $A \cup B=X$ and $A \cap B$ is contractible.