This question comes along with a lot of associated sub-questions, most of which would probably be answered by a sufficiently good introductory text. So a perfectly acceptable answer to this question would be the name of such a text. (At this point, however, I would strongly prefer a good intuitive explanation to a rigorous description of the modern theory. It would also be nice to get some picture of the historical development of the subject.)

Some sub-questions: what does the condition that d^2 = 0 means on an intuitive level? What's the intuition behind the definition of the boundary operator in simplicial homology? In what sense does homology count holes? What does this geometric picture have to do with group extensions? More generally, how does one recognize when homological ideas would be a useful way to attack a problem or further elucidate an area?

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    $\begingroup$ One sub-question you didn't mention but which interests me: Why, on a basic level, do mathematicians use cohomology more than homology, or at least talk about it more? I've developed a tentative opinion on this, but I'm always excited to hear what others have to say. $\endgroup$ Oct 15, 2009 at 20:20
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    $\begingroup$ Homology has to do with taking the free abelian group on a set, while cohomology has to do with taking the ring of functions on a set. Algebraic geometry is all about treating an arbitrary ring as the ring of functions on something, so it's not surprising that algebraic geometers care a lot about cohomology. $\endgroup$ Oct 15, 2009 at 20:51
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    $\begingroup$ Implicit in Reid's answer is the fact that cohomology has a product structure instead of just a group structure, which (for example) I believe can be used to distinguish spaces whose cohomologies are the isomorphic as <i>groups</i> but not as <i>rings</i>. $\endgroup$ Oct 15, 2009 at 23:43
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    $\begingroup$ related question:mathoverflow.net/questions/60108/… $\endgroup$ Mar 9, 2016 at 16:27
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    $\begingroup$ @HarrisonBrown IMO that is because the natural structure of a set of functions (algebra) is easier to manipulate than the natural structure of a set of spaces (co-algebra). $\endgroup$
    – Student
    Mar 20, 2019 at 14:38

10 Answers 10


Most of these links aim to give some geometric intuition for what homology does, so I'll try to briefly explain the algebraic intuition in case that's also useful.

A very common operation in algebra (e.g. algebraic combinatorics, representation theory) is to study a set by considering the free abelian group (or free k-vector space) on that set. Many sorts of questions are easier to answer in the linearized setting. Homology is basically the extension of this operation from sets to spaces. In fact, one can define the homology groups of a space as the homotopy groups of its infinite symmetric product (= free topological abelian monoid on the (pointed) space). If we work with simplicial sets rather than spaces, we see the connection to chain complexes. From a simplicial set we can form a simplicial abelian group by applying the free abelian group functor levelwise. The category of simplicial abelian groups turns out to be equivalent to the category of chain complexes of abelian groups, and the chain complex we get out is exactly the usual "simplicial chain complex" computing simplicial homology. If we started with the singular complex of a topological space, we would get out the singular chain complex of that space.

This doesn't explain why H_n measures n-dimensional "holes" in a space, but hopefully it explains somewhat why homology is important and easier to compute than homotopy (because of the "linearization" process).

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    $\begingroup$ "In fact, one can define the homology groups of a space as the homotopy groups of its infinite symmetric product (= free topological abelian monoid on the (pointed) space)." Is there a good reference for this approach? $\endgroup$ Aug 21, 2013 at 20:57
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    $\begingroup$ @DanielMcLaury Hatcher's "Algebraic Topology" has a section on the "Dold-Thom theorem" as section 4.K. Look there. $\endgroup$ May 25, 2014 at 2:07
  • $\begingroup$ One could perhaps also argue that one associates a ring to a set as the ring of functions on the set. This association would lead to cohomology rather than homology. $\endgroup$
    – Kapil
    Mar 29, 2022 at 3:00

I think the most intuitive way to look at topology is as a way to make precise the following idea. A warning: the idea by itself does not define homology, but something much scarier. Homology is what you get when you give up studying the scary but intuitive thing, and try to get something similar, but which you can calculate.

Consider a manifold. Homology is meant to count its submanifolds, up to cobordism. In other words, as out "chains of dimension $n$", take the formal sums of submanifolds of dimension $n$, where the submanifolds might have boundary. The boundary operation $\partial$ just takes the boundary. Notice that the boundary of any manifold is a manifold without boundary, so it's clear that $\partial^2=0$.

Now, the homology of this chain complex counts submanifolds without boundary; two submanifolds are considered different if they are not boundaries of the same higher-dimensional submanifold. If you think of the higher-dimensional submanifolds as a way to "move" one of the submanifolds to another, it makes sense that you might want to think of them as the "same". If a submanifold does not surround a "hole", it is the "same" as the "empty submanifold", this is a sense in which this homology counts holes.

Again, if you try to make this precise, you will run into all kinds of trouble. You'll have to define what a "submanifold" is, whether they can have self-intersections, etc. Then, you'll find that the above kind of "homology" is not really a homotopy invariant, and terribly difficult to calculate.

However, you should compare the above idea to the definition of simplicial homology. You'll see that the cycles you get in simplicial homology are similar to submanifolds, and all the wonderful algebraic machinery will show you that you can actually calculate homology of anything you'd like.

One algebraic topology book that seems to have this approach in it is Bredon's "Topology and Geometry".

The above intuition is especially useful in differential topology. There is a way to make the above idea (called cobordism theory), but you need to know how to use homology to do it. For a taste of it (that doesn't require homology), look in the last chapter of Milnor's "Topology from the Differential Viewpoint".

Three unrelated comments:

  1. Never expect any intuition from singular (co)homology, and never calculate anything with it; it is merely a tool for showing that other kinds of homology (that you actually care about) are the same and invariant under homotopy.

  2. A completely different (and very precise) way to intuitively think of cohomology is as solutions to a certain differential equation. This is the approach of the "Calculus to Cohomology" book and Bott and Tu's "Differential Forms in Algebraic Topology", and is called De Rham cohomology.

  3. Don't be surprised if there are some mistakes in any of the above; wise people - feel free to point them out and clarify!

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    $\begingroup$ I would somewhat with your first comment at the end. Part of the point of (co)homology is that it is axiomatic and completely characterized by a few basic properties--long exact sequences, homotopy invariance, and the (co)homology of a point, basically. In particular, once you have verified these properties for singular (co)homology, it is just as computable as any other formulation of (co)homology. I also don't see the intuition behind singular homology as significantly less clear than the intuition behind simplicial homology. $\endgroup$ Oct 15, 2009 at 22:17
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    $\begingroup$ I seem to recall reading somewhere that cobordism (as described in this answer) was actually Poincare's original motivation, but for a long time the world had to settle for homology because that's all that they could calculate. (The OQ asked for history as well, so I thought it worth mentioning.) $\endgroup$ Feb 18, 2016 at 15:48
  • $\begingroup$ Thank you very much for pointing out that regarding boundary of something as zero is almost the same as quotienting out cobordism relation.. they have been discerned in my mind for a long time.. now it makes perfect sense! $\endgroup$
    – Student
    Mar 20, 2019 at 14:35
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    $\begingroup$ @JoshuaGrochow About historical backgrounds: In Analysis Situs section 5, Poincaré has formulated his concept of homology $v_1+\cdots+v_p\sim0$ when $v_1,\dots,v_p$ constitutes a complete boundary of a manifold (the curve case appeared in Riemann's Theorie der Abelschen Funktionen, and according to Weibel's notes History of Homological Algebra, Betti and Riemann generalized it to higher dims), and more generally, for integer coefficients, after which he remarked that "homologies can be combined like ordinary equations". So not directly inspired by cobordism (binary relation). $\endgroup$
    – user20948
    Jan 9, 2020 at 11:43

On page 2 of this paper, while discussing Euclid's axioms of plane geometry and spatial geometry, Manin makes an extremely interesting (for me, anyway) comment:

Euclid misses a great opportunity here: if he stated the principle

“The extremity of an extremity is empty”,

he could be considered as the discoverer of the


(I've just posted a new question expanding on this.)

Now let me say a few things that are more concrete and less philosophical. The boundary operator in simplicial homology should be pretty intuitive, except for the signs. Where do the signs come from? The signs come in because, for example, we want to make sure that when we take the boundary of (say) a square, we get the same answer (namely the 4 outer edges of the square) no matter how we triangulate it. If we didn't have the signs, then different ways of triangulating the square would lead to different answers, in particular wrong answers (namely something other than the 4 outer edges of the square).

Another question might be, why simplices in the first place? Why not, say, cubes? One reason is because polygons (resp. higher-dimensional analogues) can always be chopped up into finitely many triangles (resp. higher-dimensional analogues), but they can't always be chopped up into (finitely many) squares. However, there is actually a version of homology which uses cubes instead of simplices, appropriately called cubical homology. I think the theory all works out and gives you something equivalent to simplicial homology, but it is in many ways not as nice. Though it is nicer in some ways, for example, a product of simplices is not a simplex, but a product of cubes is a cube, which makes certain proofs easier.

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    $\begingroup$ A good way of getting a handle on the basic point of homology (in addition to looking at the simplicial case) is using Z/2Z coefficients. That way, a simplex either is or isn't in a chain, and signs are meaningless. $\endgroup$ Oct 15, 2009 at 23:52
  • $\begingroup$ I came here to mention the equation "d^2 = 0", but I see someone beat me to it! $\endgroup$ Oct 16, 2009 at 0:40
  • $\begingroup$ Kevin Walker has made some interesting comments regarding related issues here: mathoverflow.net/questions/3656/… $\endgroup$ Nov 1, 2009 at 15:07
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    $\begingroup$ In his introductory book Massey develops homology with cubes instead of simplices. I think the main technical problem comes from having to factor out by some kind of degeneracy. $\endgroup$ Dec 16, 2009 at 12:55
  • $\begingroup$ I have heard of that Serre's spectral sequence construction did use "cube-like" simplices! $\endgroup$
    – Student
    Mar 20, 2019 at 14:40

On Harrison's question on why people tend to work with cohomology more than homology, one of the main reasons is that it's easier to work with. Cup products give cohomology a natural graded ring structure, and the fact that this structure is preserved by continuous maps makes it often much easier to compute cohomology than homology. For example, in the cohomology Serre spectral sequence, the differentials are derivations with respect to the product structure, and this means that by computing a small number of differentials you can usually find all of them. On the other hand, in the homology spectral sequence there are typically infinitely many possibly nontrivial differentials to compute. Besides cup products, there are other natural operations on cohomology (Steenrod operations) which are similarly computable.

Another (not unrelated) reason that cohomology can be easier to work with is that cohomology is a representable functor: H^n(X;A) is homotopy classes of maps from X to the Eilenberg-MacLane space K(A,n). By Yoneda, this means many properties of cohomology can be computed and understood by computing a single universal example. In particular, this is a deep reason why all the extra structure that makes cohomology easier to work with is computable.

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    $\begingroup$ Easier in some ways, in some approaches. The way I've come to view it is that if you approach homology/cohomology from a combinatorial direction, then homology is more natural - and if you approach it from a more geometric/analytic direction, cohomology is much more natural. Kinda like the difference between simplicial and de Rham as approaches - they tie into your mindset well from two different mindsets. $\endgroup$ Oct 15, 2009 at 20:57
  • $\begingroup$ I agree with Mikael here; my secret, partially-formed intuition is that contravariance arises from the fact that pulling back functions is so natural geometrically. But this doesn't come close to explaining all the different reasons I've heard, so I'm glad my thoughts on this are malleable. $\endgroup$ Oct 15, 2009 at 21:06
  • $\begingroup$ I should maybe also mention that I'm a big fan of homotopy theory and of combinatorics, which tells you that I don't necessarily view "easy to work with" as an advantage... :) $\endgroup$ Oct 15, 2009 at 21:20
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    $\begingroup$ As an explanation for how contravariance is related to the advantages of cohomology I gave above, note that it is almost tautological that a contravariant hom-functor Hom(-,Z) should send cofiber sequences to exact sequences (a map from X/A to Z is just a map from X to Z that is trivial on A), whereas this is not true for Hom(Z,-). This is why cohomology is representable but homology is not. $\endgroup$ Oct 15, 2009 at 21:51
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    $\begingroup$ Sheaves like to have their cohomology taken, while cosheaves (aka coefficient systems) like to have their homology taken. If you do combinatorial manipulations, you are likely to be working with coefficient systems. $\endgroup$
    – S. Carnahan
    Oct 16, 2009 at 2:58

The intuition behind the boundary operator, as well as the intuition behind d2=0 are tightly linked.

Imagine a triangle, edges labeled AB, AC and BC. What do we expect its boundary to be? The edges, in some order. If we label the vertices A,B,C and make the edge labels correspond to the vertices in the obvious manner, then we can give each edge an inherent orientation by pointing it towards the letter that occurs later in the alphabet.

And thus, going once around the triangle, we list our edges as AB + BC + CA = AB + BC - AC, using the sign to indicate when we're backing up along a directed edge.

Extending this notion of a boundary, as a bona fide geometric boundary, we notice that it ends up being linear - if we have several simplices that glue together, the resulting boundary is going to be an appropriate signed sum of the boundaries of each simplex, with the connecting edges cancelling out by the sign choice. And we can figure out a formula - the one with alternating signs and dropped vertices - that generalizes the 2d idea of a boundary to something that both looks right and works.

Now that we know the boundary of a triangle? What's the boundary of the sequence AB, BC, CA of edges?

It shouldn't take all that much to come up with the idea of "It doesn't have one". Which is, essentially, the idea behind d2=0: the boundary of a boundary should vanish, because that's what it does if we think of the geometric situation and try to formulate an intuition.

From this point on, though, a lot of algebraic topology and more importantly a lot of homological algebra takes the intuitions we build by considering easy examples, and runs with it; in homological algebra without even a resemblance of regard from the geometric origins.

  • $\begingroup$ I like this exercise even more if it includes some pathological extensions like a filled-in triangle with a spare leg (femur+tibia) hanging off. A 2-D + a 1-D object “looks ugly” and this should show up in d². $\endgroup$ Feb 16, 2020 at 4:37

Dealing with one subquestion at a time, since they require different answers.

I believe that it is very culture-dependent whether homology or cohomology seems more natural.

If you approach it all from a combinatorial approach - building spaces as triangulations, and working with decompositions of spaces - then the boundary operator is a bona fide boundary operator, and makes a lot of sense as such; and then homology falls out as the obvious thing to do with that.

If you instead approach it geometrically or analytically - with the differential forms on manifolds, or coordinate rings of functions on varieties, or a similar viewpoint as to what is the more fundamental part of your geometric situation; then cohomology - with an approach similar to de Rham cohomology over theorems linking differentials with integrals as operations on forms - is a very natural generalization of 1st year calculus material.

In addition to all this, we ALSO have that cohomology comes with a ring structure, while homology comes with a coring structure; which makes cohomology easier to work with as an algebraic object.


Of course the canonical (graduate-level?) algebraic topology text is Hatcher's, and while I've found it pretty difficult to jump into as a beginner, if you've somehow overlooked it it's probably worth checking out. (It's available for free online, I believe.)

This archived sci.math.thread (now here, hat tip to user pjfmc) helped me out with understanding what "holes" are, really. I seem to recall that John Baez has written other stuff on homology I've found very intuitive and useful, but I don't feel like digging through the TWF archives right now to find it -- I may edit this later.

Finally, if you have access to a copy of the PCM, Burt Totaro's article on algebraic topology is quite well-written and intuitive, although there's probably more material on homotopy than (co)homology there. His references include Hatcher as well as Armstrong's Basic Topology at a lower level (which I own, and while I quite like it in general, I'm not a fan of the chapter on homology), Bott and Tu's Differential Forms in Algebraic Topology and Milnor's stuff.

  • $\begingroup$ The link to the sci.math thread does (no longer?) work, can a valid pointer to the document be restored? $\endgroup$
    – Dilaton
    Apr 12, 2018 at 13:22
  • $\begingroup$ @Dilaton unfortunately, Harrison does not seem to be active anymore (Last seen Sep 17 '12 at 23:55) :( $\endgroup$
    – byk7
    May 13, 2019 at 11:26
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    $\begingroup$ I was able to track down the archived sci.math thread mentioned by Harrison Brown. I think the discussion is helpful. web.archive.org/web/20020609160528/http://www.math.niu.edu/… $\endgroup$
    – pjfmc
    Mar 28, 2022 at 21:17

This is a very broad question, so I'm just going to suggest some references. It seems that the most popular introductory book on algebraic topology is Hatcher's book, which is available on his webpage


Another good source with a different point of view is Bott-Tu's book, which deals with deRham cohomology and thus might be easier to follow. The wikipedia article at


is also pretty good. Finally, for the history of the subject you can't beat Dieudonne's book "A History of Algebraic and Differential Topology: 1900-1960.". It's a bit too much in the very beginning, but once you've learned to solve the "finger exercises" in the intro textbooks it will give you a lot of intuition for the subject.


The connection to group extensions is much more subtle, and not entirely easy to tie in with all these much more fundamental topological questions.

As I mentioned in another answer, much of homological algebra is about using the techniques from algebraic topology, but in situations where the geometrical underpinnings vanish from sight.

Basically, it turns out that the process of going from, say, a triangulated space to its homology groups is by building chain complexes. And finding chain complexes that replace the space, retaining some of its properties.

So, for studying group extensions, it turns out that doing something similar - constructed either by building a topological space out of the group, or by building chain complexes directly through a somewhat ad hoc construction - we can say things about the group. But studyin group extensions through group cohomology is much easier to understand by approaching it with studying how certain functors deal with short exact sequences, than by trying to frame it in terms of geometry and geometrical topology.

The connection, really, appears in that in both cases, the way to find an answer is through replacing the original object with a chain complex, and then doing Stuff, including finding homology groups, to that chain complex.


try this:

From calculus to cohomology

  • $\begingroup$ I had good luck learning about DeRham cohomology from this text, as well as the aforementioned Bott and Tu. $\endgroup$ Dec 16, 2009 at 20:48

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