# What is the process of computing homology? [closed]

Suppose you have a torus. You triangulate it. Suppose you want to compute $H_{1}(X)$. If you let $z$ be a $1$-cycle, why do you change it by a multiple of a triangle to get a homologous cycle? What is the algorithm? Why do we do this?

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## closed as too localized by Anweshi, Robin Chapman, Loop Space, Qiaochu Yuan, Ryan BudneyAug 4 '10 at 19:41

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Have you looked at en.wikipedia.org/wiki/Homology_%28mathematics%29 ? – S. Carnahan Aug 4 '10 at 18:39
Ross, please take a look at some expositions of homology. The introduction in Hatcher's book math.cornell.edu/~hatcher/AT/ATpage.html gives you a sense for the type of geometric issues one is trying to formalize. Initially the idea of a homology goes back to Poincare. Dieudonne has a nice summary of Poincare's ideas in his "History of Algebraic and Differential Topology", although it might be best to put that off until you feel a little more oriented towards the homology idea. – Ryan Budney Aug 4 '10 at 19:38

I'm not completely sure if I understand the question, but I will at least try to answer the "why we do this" part. I think you're asking, "Why do we look at cycles modulo boundaries instead of just cycles?" and the answer to this is pretty simple: boundaries are uninteresting. If some $n$-cycle is the boundary of an $(n+1)$-simplex, then the boundary can be "pulled across" the simplex to no longer represent a "loop." In otherwords; boundaries don't represent true holes. We care about cycles that surround a hole, so that's why we look at cycles modulo boundaries.

I'm not sure what you mean by "what is the algorithm." If you mean, "how do you compute homology?", then you can look that up in any algebraic topology book or wikipedia.

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But when I compute $H_{1}(X)$ for the torus I ask the following: What are the number of $1$-cycles that are not $1$-boundaries? It is the two circles that one cuts to make the torus into a rectangle. So $H_{1}(X) = \mathbb{Z} \times \mathbb{Z}$. Is this an okay way of thinking about homology? – Ross Aug 4 '10 at 18:34
Well there are lots of other 1-cycles that are not boundaries, since curves can loop around both "holes" in the torus in complicated ways. The point is that these are homologous to linear combinations of the two classes you suggest. – Jack Huizenga Aug 4 '10 at 21:03

One way of thinking about this is that homology gives a way to almost answer the questions you would try to answer by homotopy, but in a way that allows you to use linear algebra to get the answer. You don't get all the way - which is why homology and homotopy are actually different entities - but anything that would have been identified by homotopy certainly will be identified by homology.

Key in this process is the transferral from topological arguments to algebraic arguments: we would be interested in, say, non-trivial loops under the equivalence relation given by homotopies of functions, but that's topological and potentially hard. Instead, we find correspondences in algebra to what non-trivial and loop means.

Thus, we triangulate the space to get a simplicial approximation of the space, and instead of looking at bona fide loops, we look at cycles. These are like loops in that neither have non-trivial boundaries - the loops don't have non-trivial boundaries because of their geometry, the cycles don't by the way we define cycles: namely as things that are in the kernel of the boundary operator. (drawing pictures of simplicial 1-cycles at this stage helps understand why these definitions give us the geometric results we want to approximate by the algebraic approach)

Similarily, we want to capture homotopic loops as somehow the same. Specifically, this means we should be able to wiggle our loops around as long as they stay in the space. Once we go to the simplicial side, the wiggles have to jump across an entire simplex in one go, changing an edge to the other two edges of a triangle, say. The upshot of all this is that in order for homotopic things to be identified in homology, we need to identify anything that differs by the action of pulling a cycle across a higher-dimensional simplex.

Which is, once the pictures are drawn and the arguments you'd find in any introduction to algebraic topology are made, the same as taking the factor group by the image of the boundary operator.

In the end, to compute $H_1(X)$, what you really compute is a complete enumeration of all possible cycles, under the condition that things you get by wiggling a cycle a little bit should be considered the same.

Now, the next question would be Why do we care about homotopic functions? which is a very different question that I won't start answering here.

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