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First of all I don't know much about homological algebra, and algebraic topology I just took a class using Kinsey book (Topology of Surfaces) which is undergraduate math class and I learned what is homology group and how to calculate it from a complex.

This frist defines a directed cell complex and then generate $k$-chain group from $k$-skeleton. After that considering chain complex, with boundary operation, kinsey defines homology group.

However, Munkres and Hatcher and other algebraic topology book does not follow this way. Every text I have generate $k$-chain group from Simplicial complex not, directed cell complex. (Hatcher uses $\Delta$ complex but every $Delta$ complex can be subdivided into simplical so this does not matter).

I want to know how Kinsey's process makes sense. Please Help.

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    $\begingroup$ A chain complex is just a sequence of abelian groups $C_k$ and boundary operators $\partial_k: C_k \to C_{k-1}$ with $\partial^2 = 0$. The homology of a chain complex is $H_k = \text{ker}(\partial_k) / \text{im}(\partial_{k+1})$. The difference between the presentations you saw is how they defined $C_k$ and $\partial_k$, not the definition of homology of a chain complex. I have no idea what a directed cell complex is, but I imagine that when it gives you the same space as a simplicial complex, Kinsey's definition and Hatcher's will give you isomorphic homology groups. $\endgroup$
    – user98602
    Jul 27 '15 at 17:32
  • $\begingroup$ I also have no idea what a directed cell complex is, but using simplicial methods are good for giving proper signs or orientations to the boundary maps, which keeps things simple when defining the singular homology for arbitrary spaces. $\endgroup$
    – JWL
    Jul 27 '15 at 17:37
  • $\begingroup$ I think directed cell complex is CW-complex with orientations on every cell. What I want to know is that it reasonable to generate abelian groups with k-skeleton from CW-complex, not from delta complex. $\endgroup$
    – Allen Cho
    Jul 27 '15 at 17:38
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    $\begingroup$ @AllenCho Maybe you can have a look at Cellular Homology. Hatcher defines singular homology with simplicially, then proves the equivalence of singular homology and cellular homology in the book. $\endgroup$
    – JWL
    Jul 27 '15 at 17:44
  • $\begingroup$ A CW complex automatically has orientations on every cell by the way it's defined. In any case, yes. The standard way of doing this is usually known as cellular homology. It's isomorphic to simplicial homology for simplicial complexes. $\endgroup$
    – user98602
    Jul 27 '15 at 17:45
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It sounds as if you're confused about the difference between singular/simplicial homology and cellular homology. If you look at the homology chapter of Hatcher's book again, you'll see that these subjects are covered in some detail.

The relevant details are as follows:

  • The homology of a space is a deep property of that space that is hidden behind a frosted glass wall. That is to say, it should be considered as an extremely fundamental topoogical invariant, but not one which is readily accessible to us. This means that it is difficult to talk about the meaning of the homology groups - there is a nice interpretation in terms of the $n$-dimensional 'holes' in the surface, but that breaks down once we start encountering spaces whose homology groups exhibit non-trivial torsion. This is why we often have to do a lot of messy work in order to get anywhere with homology. Simplicial homology is not a nice piece of mathematics, and it's not obvious why we care about it at all and why we are prepared to invest so much effort studying it. The reason is that it gives us a way to reach through the glass and grasp the homology of the space.

  • The nice thing is, once we have got a foothold on the homology groups of a few spaces, we can apply various rules again and again to compute many more homology groups, and the whole subject opens up to us.

  • Pretty soon, we can identify exactly which tools we are using again and again. This leads into the idea of defining a 'homology theory' axiomatically: we just specify that it is a rule that takes a topological space and spits out a sequence of groups in such a way that all our useful tools work. These tools - homotopy invariance, excision, the dimension rule, additivity and the LES of a pair - are known as the Eilenberg-Steenrod axioms.

  • We then see a reason for computing simplicial homology in the first place - it gives us a model of the Eilenberg-Steenrod axioms. Simplicial homology satisfies the Eilenberg-Steenrod axioms, and so we are guaranteed that A) It is a homotopy invariant and B) we can compute it in a nice way using the other tools. Homotopy invariants that are easy to compute are extremely useful to algebraic topologists!

  • Where does cellular homology come into this? Well, the other cool thing about homology theories is that they tend to agree. Remember what I said about homology being a deep but somehow 'hidden' feature of a space? Well, simplicial homology is one way to reach into the heart of a space and extract this important invariant. But there are a number of other ways, and it turns out that cellular homology (computing homology from the chain complex of $n$-cells) is one such way. Moreover, cellular homology is much easier to compute than simplicial homology.

  • The problem is, in order to prove anything about cellular homology, we need to have the theory of homology set up already. So we need some homology theory already set up that satisfies the Eilenberg-Steenrod axioms. That's where simplicial homology (or, more commonly, its cousin singular homology) comes into play. We need to define it first, but once we've used that to set up the theory, we can use it to prove that cellular homology A) satisfies the Eilenberg-Steenrod axioms and B) agrees with our previous definition of homology, and then we can use the (easier) cellular homology to compute homology groups for cell complexes from then on.

Important point (suggested by Mike Miller in the comments): The definition of cellular homology involves picking a choice of cell decomposition for our space. A cell complex may admit many different cell decompositions. How do we know that the cellular homology we get is independent of our choice of decomposition? Why, because we know it is isomorphic to singular homology, which is a topological(/homotopy/weak homotopy) invariant.

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  • $\begingroup$ I don't know how to thank you about your careful response. Thank you very very very much !!! I have one more question, I heard that there exists a manifold that does not have simplicial complex, but every compact mainfold have appropriate CW-complex. Is cellular homology well calculated even for this kind of weird manifold? $\endgroup$
    – Allen Cho
    Jul 27 '15 at 17:55
  • $\begingroup$ @AllenCho Nice question! The answer is YES, and the reason is singular homology. Singular homology is quite similar to simplicial homology, but it can be calculated for every topological space. That eliminates any difficulties arising from non-triangulable manifolds. If simplicial homology were our only tool, you are right that this would be a problem. $\endgroup$ Jul 27 '15 at 18:00
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    $\begingroup$ @AllenCho: Every smooth manifold is homeomorphic to a simplicial complex. Outside of dimension 4, every topological manifold is homeomorphic to a CW complex; it's open whether or not this is true in dimension 4. However, every manifold is homotopy equivalent to a CW complex. In any case, cellular homology is defined for CW complexes. You don't just need a space, you need a CW structure on that space. This turns out to be isomorphic to something that doesn't require a CW or simplicial structure at all, called singular homology. $\endgroup$
    – user98602
    Jul 27 '15 at 18:01
  • $\begingroup$ Please let me have one more question, You said axiomatic approach of homology theory is to make sth to satisfy Eilenberg-Steenrod Axioms. Is homology group unique algebraic object that satisfying this axiom? $\endgroup$
    – Allen Cho
    Jul 27 '15 at 18:01
  • $\begingroup$ On finite CW complexes the Eilenberg-Steenrod axioms completely specify the homology theory. I'm sure you can find different homology theories that satisfy them, but they would automatically agree on finite complexes. $\endgroup$
    – user98602
    Jul 27 '15 at 18:03

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