# Homology - why is a cycle a boundary?

I have a question about the basic idea of singular homology. My question is best expressed in context, so consider the 1-dimensional homology group of the real line $H_1(\mathbb{R})$. This group is zero because the real line is homotopy equivalent to a point. The chain group $C_1(\mathbb{R})$ contains all finite formal linear combinations of continuous maps from the interval $[0,1]$ into $\mathbb{R}$. One such map (call it $\mu$) maps the interval along some path that begins and ends at zero. (For my purposes it doesn't matter how exactly.) This map is a cycle, i.e. is contained in the kernel of $\partial_1:C_1 \rightarrow C_0$, because it begins and ends at the same point. It must be that it is also a boundary, i.e. contained in the image of $\partial_2:C_2 \rightarrow C_1$, because otherwise it would represent a nonzero homology class in $H_1$. My question is about exactly how and why it is a boundary.

I have an intuitive understanding of why it is a boundary that does not seem to work when I translate it into formal language, and a formal way to show it is a boundary that does not seem to capture the heart of the intuition. My reference on the formal definitions is Allen Hatcher's Algebraic Topology.

Intuitively, $\mu$ maps $[0,1]$ to a loop and then smooshes it into the real line (i.e. $\mu$ factors through $S^1$). The map from the loop to the line could be extended to a disc without losing continuity, since the whole thing gets smooshed anyway. A triangle could be mapped homeomorphically to the disc, and this would give us a map $\zeta: \Delta^2 \rightarrow \mathbb{R}$ of which, intuitively anyway, $\mu$ is the boundary. However, formally, $\partial_2 (\zeta)$ is the formal sum of the restriction of $\zeta$ to each of its edges; it is thus a formal sum of three maps from the interval to the real line, and thus is not (formally) equal to $\mu$.

Formally, I can define a map $\alpha : \Delta^2 \rightarrow \mathbb{R}$ from a triangle to the real line that does have $\mu$ as a boundary, but I am very unsatisfied with this construction because it involves details that feel essentially extrinsic to the intuition above. Let the vertices of $\Delta^2$ be labeled 0, 1, 2. Map $\Delta^2$ to a disc in the following way: map vertex 0 to the center of the disc; the edges $[0,1]$ and $[0,2]$ to a radius in the same way (so that the restrictions of $\alpha$ to the two edges are equal); the edge $[1,2]$ around the circumference; and extend the map to the interior of the triangle in the obvious way. Then map the disc to the real line as above; the restriction to the circumference is $\mu$. Now, the boundary map $\partial_2$ by definition maps $\alpha$ to $\alpha |_{[0,1]} +\alpha |_{[1,2]}-\alpha |_{[0,2]}$. But $\alpha |_{[0,1]}$ and $\alpha |_{[0,2]}$ are equal and $\alpha |_{[1,2]}$ is equal to $\mu$, so $\partial_2(\alpha)=\mu$.

My question is this: is it correct that the intuitive construction of $\zeta$ does not provide an element of $C_2$ with $\mu$ as a boundary? Is it correct that in order to get $\mu$ as a boundary one must use a construction like that of $\alpha$ above? If so, is the intuition that $\mu$ is a boundary because it is a loop that can be extended to a disc before smooshing wrong? Does the fact that $\mu$ is a boundary really hang on the sign convention in the definition of $\partial_2$? If so, can you give me a reason for why this sign convention works to guarantee that such a construction will always exist when a cycle "seems like it should be" a boundary?

EDIT:

EDIT (7/12):

Thanks for all the help everyone. My immediate acute sense of cognitive dissonance has been addressed, so I'm marking the question answered. I have some residual sense of not getting the whole picture, but expect this to resolve itself with slow processing of more theorems (like homotopy invariance of homology, and the Hurewicz map, thank you Matt E and Dan Ramras).

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You might want to take a look at the Hurewicz map, which is a homomorphism from the homotopy groups of a space to the homology groups. This means, among other things, that nullhomotopic maps from a sphere to a space yield nullhomologous cycles. Of course this just begs the question to some extent, but if you find a proof of the Hurewicz theorem that you like, it may give you some further intuition. (Finding such a proof may be a challenge!) – Dan Ramras Jul 10 '11 at 1:11

Your intuition is correct, I think. I also had the experience, when first learning this material, of wanting to understand homologies explicitly in the way that you are trying to, so I encourage you to pursue your attempt to match intuition with formal definitions.

The basic problem you observed is that often, at a technical level, one has to produce formal sums of cycles, while when thinking intuitively, one doesn't normally generate these formal sums in one's imagination. The way to reconcile this is to prove the following:

If $\alpha:[0,1] \to X$ and $\beta: [0,1] \to X$ are two 1-simplices (in any target space $X$) and $\gamma:[0,1] \to X$ is the sum of $\alpha$ and $\beta$ in the sense of addition in the fundamental group, then there is a homology between $\alpha + \beta$ (formal sum) and $\gamma$. This is easily checked, so I leave it as an exercise. (In a complete treatment of singular homology, it would appear as part of the verification of homotopy invariance, probably in some implicit manner. It is also closely related to Dylan Wilson's suggestion about verifying that homotopic cycles are homologous.) Once you've done this, you'll have more confidence that various intuitive pictures do indeed match with the more formally correct treatment.

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+1 This is extremely helpful - if I could vote it up twice I would. This kind of result is just what I was hoping for. (It does hang on the sign convention in the definition of $\partial_1$, but is making me feel like this convention is more meaningful...) Is there an analogous result in dimension 2 and higher dimensions? (I am having trouble imagining it in the even dimensions - see my comment on MartianInvader's answer.) Thanks to your tip I'll consult Hatcher's proof of homotopy invariance. – Ben Blum-Smith Jul 9 '11 at 14:48

Your construction works, but I think it's a little more complicated than it needs to be. In particular, you can just define a map from a simplex directly, without having to go through a disc or anything.

I'd do it by taking a triangle (again, let's call the vertices 0,1,2) and mapping [0,1] and [0,2] via the map that is constantly zero, and mapping [1,2] to the loop $\mu$. This can clearly be extended to the interior of the triangle ($\mathbb{R}$ is simply connected after all).

When you take the boundary, the [0,1] and [0,2] edges map to the same constant map, and thus cancel each out, so you're left with only $\mu$ in the boundary.

I agree that it's a little clunky that you need a canceling trick to turn three edges into one (and you'd need to do something similar for a 1-cycle involving two pieces), but it does end up working, and the same trick shows that any nullhomotopic loop is a boundary.

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+1 Very helpful. I still feel very unsatisfied but maybe I have to live with this. Taking things to dimension 2, consider a map $\mu$ from $\Delta^2$ to a sphere in $X=\mathbb{R}^3$ that maps the boundary of the triangle to a point. Intuitively, $\mu$ "should be" a boundary because it maps the triangle to a sphere that is filled in by a ball; equivalently, it is nullhomotopic in $X$. But how do I formally realize it as a boundary? Since $\Delta^3$'s have 4 faces, before cancellation there are an even number of maps in the boundary of any chain in $C_3$. How will they cancel to leave $\mu$? – Ben Blum-Smith Jul 8 '11 at 23:36
@Ben Blum-Smith: The example of a sphere you gave isn't actually a boundary. It looks like one because it "folds over" itself, to include the entire sphere, but it's actually just a single triangle, which doesn't bound a volume. In fact, it's not even a cycle - applying the boundary operator to it should give you three constant maps from 1-simplices, two of which cancel, and you don't end up with zero. To make it a true cycle/boundary, you have to attach another triangle. You may even be able to turn this into a proof that all true boundaries use an even number of triangles. – MartianInvader Jul 11 '11 at 18:29
Oh, got it. Yes, it seems to follow immediately that a chain in $C_2$ is not a cycle unless it has an even number of triangles, so the cycle I was really going for contains one triangle mapped to the sphere, mapping the whole boundary to a basepoint, and then another (negative) triangle mapped constantly to the basepoint; and I can see how to realize this cycle as a boundary. Thanks again, all these answers have been very clarifying. – Ben Blum-Smith Jul 12 '11 at 15:35

Remember that $C_2X$ does not just consist of all maps $\sigma: \Delta^2 \rightarrow X$ but also all formal sums of these. In particular, consider a map of the unit square that realizes a nullhomotopy of $\mu$. Divide the unit square into two triangles, label the vertices, orient the edges properly, and interpret the nullhomotopy as a sum of two different maps $\Delta^2 \rightarrow X$ (one might have a minus sign, actually). The boundary of this chain should be $\mu$, if all goes well.

I think, actually, this type of thing should work more generally to show that if two paths are homotopic, then they are homologous (i.e. differ by a boundary.)

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Thanks Dylan - I'm actually still confused by this, but I think that my confusion is at the heart of the matter, so I feel we're getting somewhere. – Ben Blum-Smith Jul 8 '11 at 20:19
In your construction, let's call the map from the square to the line, $\Mu$. It seems to me that the restriction of $\Mu$ to "the other three" edges of the square will be a constant map to the base point. (The restriction to the fourth side is $\mu$.) Then $\partial(\Mu)$ is a sum of six maps from the unit interval into $\mathbb{R}$: three are constant to the base point (say 0), 2 are equal from the square's diagonal to some path, and one is $\mu$. If I do the orientations just right, maybe I can get the 2 from the diagonal and 2 of the three constant maps to cancel. – Ben Blum-Smith Jul 8 '11 at 20:29
But amn't I still left with at least one of the constant maps in addition to $\mu$? I figure that to get rid of it I can add or subtract another map $\Nu: \Delta^2 \rightarrow \mathbb{R}$ to the original chain in $C_2$ which is constant to the base point, and then its boundary will cancel out the extra constant map, but the question is, <i>is all this finicky, unnatural-feeling-to-me sign stuff really needed?</i> And if so, how do I know it will always be available? (Preferably, how can I <i>understand intuitively</i> that it will always be available, but I know this is a tall order.) – Ben Blum-Smith Jul 8 '11 at 20:38
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