Terminology question: "boundary map" in homology theory

Question

In homology theory the same name and symbol is used to describe two different maps:

1: the boundary maps $\partial_n: C_n(X) \to C_{n-1}(X)$ appearing in a chain complex of (say) singular n-simplices.

2: given a pair (X,A) the boundary maps $\partial_n: H_n(X,A) \to H_{n-1}(A).$

Is there a good reason for this? To my eyes, these maps seem quite unrelated: for instance, the second map takes (equivalence classes of) cycles to (equivalence classes of) cycles, while the first map annihilates all cycles!

Answer 1

This is just an elaboration on PVALs comment.

Given a short exact sequence $0 \to A \to B \to C \to 0$ of chain complexes, there is an induced long exact sequence in homology. This is constructed out what is known as the snake lemma: https://en.wikipedia.org/wiki/Snake_lemma

If you look in the construction of the boundary map, you will see that a key step is to use the differential of $B$. This can be interpreted topologically.

First lets just recall the construction of the relative chain complex short exact sequence. The idea is that the inclusion $A \to X$ induces an inclusion of (singular) chain complexes, and one can formally take the quotient to get the relative chain complex. (The quotient is also free, since a simplex is killed in the quotient iff it lied entirely in $A$, so we create no torsion after this quotient.)

Now we start with some representative $\alpha$ of an element in the relative homology group $H_n(X,A)$, which is some chain in $X$ whose boundary lies in $A$. $\partial \alpha$ is some chain lying in $A$, and since it was a boundary of a chain in $X$, we know that it is a cycle. But it may not be a boundary of some chain lying in $A$, hence could be some interesting homology class in $H_{n-1}(A)$.

It is useful to imagine the disc $D = X$ with boundary $S^1 = A$. The disc is a relative homology class in $H_2(X,A)$. It's boundary is a homology class in $H_1(A)$.

The long exact sequence asserts that in general a complete set of representatives for the homology classes in the kernel of the map $H_{n-1}(A) \to H_{n-1}(X)$ are obtained this way. But this really makes sense, since some $\beta$ being in the kernel of that map means that there appeared some chain $\alpha$ in $X$ filling in the holes of $\beta$, i.e. it became a boundary, or symbolically $\partial(\alpha) = \beta$. But now $\alpha$ is a chain in $C_n(X)$ whose boundary lies in $A$, hence it represents some relative cycle in $H_n(X,A)$, and its image under the map described before is exactly $\beta$.

Hope that is helpful.

Answer 2

As both PVAL and AreaMan have suggested, the two boundary maps really are doing the same thing. In both cases, $\partial_n$ maps an $n$-dimensional cycle to zero - but in the second case, "zero" really means "in $A$." To elaborate:

As you mentioned, $\partial_n:C_n(X)\to C_{n-1}(X)$ annihilates all cycles, i.e. if $c\in C_n(X)$ is a cycle then $\partial_n(c)=0$. Intuitively this is because, well, a cycle has no boundary! (As a back-pocket example, I like to think of a 1-chain as the unit interval [which has boundary, the two endpoints] and a 1-cycle as a circle or some type of loop [which has no boundary, since the two endpoints are identified]. So intuitively $\partial_1$ must map the latter to $0$. Similarly for higher dimensional things.)

In the second case, $\partial_n:H_n(X,A)\to H_{n-1}(A)$ still maps $n$-cycles to "zero" - but here, in the quotient, zero is really $A$! (Just like in a quotient group $G/H$, when we say "$g\in G/H$ is zero" what we really mean is "$g\in H$.") Intuitively, this is because a cycle in $H_n(X,A)$ is a really just a chain in $X$ whose boundary lives in $A$. And in fact, it turns out that $H_n(X,A)\cong \tilde H_n(X/A)$ for all $n$ (see, for ex., Hatcher's Proposition 2.22). So ultimately we're collapsing all of $A$ to a point, and in the process, all those chains in $X$ whose boundaries lived in $A$ become bona fide cycles which really do have zero boundary.

Of course, there is some equivalence class business going on, but I hope this helps at least intuitively!