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.

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    $\begingroup$ The Mayer-Vietoris sequence is a good tool for spaces like this. In practice, you use the MV sequence by writing it down, filling in what you know (e.g. the homology groups for any spheres in the sequence, the homology groups of 1-complexes which are easy to compute, zero homology groups, etc). In 90% of the introductory algebraic topology examples, this will be enough to let you compute the remaining unknown groups in the sequence without detailed knowledge of what the boundary maps actually are, just by using the properties of exact sequences. $\endgroup$ Jul 5, 2018 at 13:21

2 Answers 2


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.


Since you know how to compute its fundamental group, the first homology group is the abelianization of the fundamental group. The zero homology groups is isomorphic to $\mathbb{Z} $ since it is connected. And top homology group is zero since it is not compact.

Top homology of an oriented, compact, connected smooth manifold with boundary

  • $\begingroup$ You're sure that the removed points do not ruin the second homology group? $\endgroup$
    – Arthur
    Jul 5, 2018 at 12:59
  • $\begingroup$ @Arthur removing finitely many points from a torus will make its $H_2$ vansih. $\endgroup$ Jul 5, 2018 at 13:05
  • $\begingroup$ Given that the OP's question is about the basics of how to even use LES's to compute homology, the question you linked seems inappropriately high level (the answers to that question reference Poincare duality and cohomology--both of which the OP is not likely to have studied yet). As such, this answer seems incomplete at best and likely to intimidate rather than inform the OP at worst. $\endgroup$ Jul 5, 2018 at 13:36

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