# What is the geometric motivation for quotienting out boundaries in homology?

I know that being contractible is strictly stronger than being homotopically trivial, see e.g. this question. Now, trying to understand p. 100 of Hatcher's Algebraic Topology, the intuitive reason we factor out $n-$cycles which enclose $n+1$-cycles might be because they are homotopically trivial.

The first example on p.100 suggests that $1$-cycles that are the boundaries of $2-$cycles can be contracted to a point, and thus are not only homotopically trivial but also contractible.

How generally applicable is this way of thinking? Is "we factor out $n$-cycles that enclose $n+1$-cycles to account for the fact that they are contractible" valid geometric intuition for the definition of homology groups? (Or at least for cellular or singular homology?)

The idea seems to be along the lines that "boundaries of (n+1)-cycles correspond geometrically to the absence of holes, and thus we don't want to count them" and "n-dimensional cycles that are not boundaries of (n+1)-cycles enclose n-dimensional holes". Is this way of thinking correct?

If not contractibility, is there any formal way we can express the fact that "boundaries of (n+1)-cycles correspond to the absence of holes", besides per fiat?

The graph $X_1$:

Let us now enlarge the preceding graph $X_1$ by attaching a 2-cell $A$ along the cycle $a-b$, producing a 2-dimensional cell-complex $X_2$. If we think of the 2-cell $A$ as being oriented clockwise, then we can regard its boundary as the cycle $a-b$. This cycle is now homotopically trivial since we can contract it to a point by sliding over $A$. In other words, it no longer encloses a hole in $X_2$. This suggests that we form a quotient of the group of cycles in the preceding examples by factoring out the subgroup generated by $a-b$. In this quotient the cycles $a-c$ and $b-c$, for example, become equivalent, consistent with the fact that they are homotopic in $X_2$.

$X_2$:

Algebraically we can define now a pair of homomorphisms $C_2 \overset{\partial_2}{\to} C_1 \overset{\partial_1}{\rightarrow} C_0$ where $C_2$ is the infinite cyclic group generated by $A$ and $\partial_2(A)=a-b$. The map $\partial_1$ is the boundary homomorphism in the previous example. The quotient group we are interested in is Ker$\partial_1$/Im $\partial_2$, the kernel of $\partial_1$ module the image of $\partial_2$, or in other words, the 1-dimensional cycles modulo those that are boundaries [of 2-dimensional cycles], the multiples of $a-b$. The quotient group is the homology group $H_1(X_2)$... In the present example $H_1(X_2)$ is free abelian on two generators, $b-c$ and $c-d$, expressing the geometric fact that by filling in the 2-cell $A$ we have reduced the number of 'holes' in our space from three to two.

My question being: how do we express the motivation for modding/quotienting/factoring out the boundaries in topological terms? In geometric terms?

• You seem to be putting words in Hatcher's mouth. He does not say (or write) that being homotopically trivial is the intuitive reason we factor out an $n$-cycle which bounds an $n+1$ cycle. And it's a good thing too, because that would be false. It is very common to have an $n$-cycle that bounds an $n+1$ cycle but that is not "homotopically trivial" in any sense of the word. – Lee Mosher Nov 12 '16 at 17:46
• @LeeMosher Re-reading how I wrote it, I agree, I was putting words in his mouth -- I tried to edit the question to reflect that that was just my attempt at interpreting the quoted passage. – Chill2Macht Nov 12 '16 at 18:17
• I've added a few thoughts to my answer. – Lee Mosher Nov 12 '16 at 19:37

This is just one example. In this example, there is only one 2-cell, and its boundary cycle $a-b$ is what is contractible.

To continue the quote, "...we form a quotient ... by factoring out the subgroup generated by $a-b$".

In other examples, you will have a larger number of generating elements that are being quotiented out. Those generating elements will generate, by the laws of algebra, an entire subgroup. That subgroup is called the "boundary subgroup". In the basic construction of homology, perhaps the generating elements of the boundary subgroup are contractible in some sense. But the group of cycles is an abelian group, and the formal algebra of abelian groups is forcing many other cycles to be quotiented out. In fact, the boundary group is nothing more nor less than the group of linear combination of the generating elements. So while certain generating elements have been singled out for special behavior, e.g. for being "contractible" in some sense, you can not make any deductions about the "contractibility" of a general element of the boundary group.

ADDED: Here's a few more thoughts, regarding what is intuitive in homology. Homology is abelian, unlike the fundamental group. This makes it computable and comparable in great generality, unlike the fundamental group. This whole business of "cycles" and "boundaries" is kind of what is forced upon you when you start with the fundamental group properties of "paths" and "homotopies" and attempt to abelianize them. In fact, one test of homology theory is that if you take the fundamental group, and abelianize it (a purely algebraic operation for converting arbitrary groups into abelian groups), the result turns out to be the 1st homology group.

• So (for dimension $n=1$ only) a boundary is kind of like path/loop that encloses a homotopy? And then geometric "presence of a hole" can be thought of topologically as the "infeasibility of a connecting homotopy"? Obviously this is more difficult to generalize for $n>1$. Except everything is abelianized, so we have cycles instead of loops, and then that is why $H_1(X)$ is the abelianization of the fundamental group. – Chill2Macht Nov 12 '16 at 20:21
• No, you cannot guarantee that a boundary is like a path/loop that encloses a homotopy. This is generally true for the generators, but no guarantees for other boundaries in dimension $1$. What is true is that a boundary is a "kind of like path/loop" that encloses a 2-chain. – Lee Mosher Nov 12 '16 at 20:32

1. This looks like it is wrong -- i.e. being a boundary in some chain group does not imply being contractible, because being a boundary implies having trivial homology, and it is not necessarily the case that trivial homology implies being contractible -- another necessary condition is having a trivial fundamental group. Note that in most cases the Hurewicz theorem only gives a homomorphism from the fundamental group to the first homology group, not an isomorphism, so the triviality of the first homology group need not imply the triviality of the first fundamental group.

It does look like Poincare's intuition when first developing homology theory may have been that all boundaries should be contractible, but then he found a counterexample in the Poincare homology sphere. For all of the details from which this answer was based, see this related question: Do trivial homology groups imply contractibility of a compact polyhedron?

2. This Wikipedia article also seems to imply that sufficient (necessary?) conditions for a boundary to be contractible are that it be a CW complex with trivial fundamental group. In general it seems that vanishing homology groups (corresponding to being a boundary?) do not necessarily imply that the space is contractible, even for CW complexes, see for example here.

Thus saying that boundaries are contractible may be nice intuition for CW complexes in special cases, but it seems also to be important to recognize that it can't always be true.

Even for CW complexes, including manifolds with boundary (I think) there are counterexamples in the other direction of contractible spaces that are not boundaries (assuming that the boundaries do just correspond to disks $D^n$), at least if I am interpreting the answers to this MathOverflow question correctly (which is to some extent doubtful).

Here is an advanced category-theoretic treatment of spaces with trivial homology groups (which I don't understand). It might be relevant or of interest to those reading.