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And so this week, our algebraic topology class starts with relative homology groups. But there are some (REALLY) basic parts of the definition of the relative homology group that I don't understand why...

Our class is currently using Hatcher's Algebraic Topology.

Given that $A$ is a subspace of a topological space $X$.

Hatcher defines the chain group $C_n(X,A)=C_n(X)/C_n(A)$.

I understand that $C_n(A) \subset C_n(X)$.

1) Why is $C_n(A)$ even a subgroup of $C_n(X)$ in the first place?

2) I understand why the boundary map $\partial$ takes $C_n(A)$ to $C_{n-1}(A)$, but why does it induce a quotient boundary map from $C_n(X,A)$ to $C_{n-1}(X,A)$?

I apologize if it is really simple and probably obvious to most but I really can't see why.

ireallydonknow
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1 Answers1

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Ad 1), since $A\subset X$, every chain in $A$ is also a chain in $X$. Then $C_n(A)$ is the subgroup of chains such that the image of every singular simplex happens to lie in $A$. The empty chain $0$ is a chain in $A$, since it contains no (singular) simplex whose image is not contained in $A$. If $c_1,c_2 \in C_n(A)$, then every simplex in $c_1+c_2$ lies in $A$, hence $c_1+c_2\in A$, and finally, for $c\in C_n(A)$, the chain $-c$ also consists only of simplices in $A$, hence $-c \in C_n(A)$. So $C_n(A)$ is a nonempty subset of $C_n(X)$ that is closed under the group operations, hence a subgroup.

Ad 2), we have the composition

$$\varphi = \pi \circ \partial \colon C_n(X) \to C_{n-1}(X) \to C_{n-1}(X,A)$$

of the boundary map and the canonical projection. That is a homomorphism, and

$$C_n(A) \subset \ker \varphi$$

since $\partial C_n(A) \subset C_{n-1}(A) = \ker \pi$. Thus we have an induced homomorphism

$$\overline{\varphi} \colon C_n(X)/C_n(A) \to C_{n-1}(X,A).$$

This induced homomorphism is the quotient boundary map.

Daniel Fischer
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  • I don't really understand how that implies that $C_n(A)$ is a subgroup of $C_n(X)$. For the second part, is the induced homomorphism implied from the Factor Theorem? – ireallydonknow Mar 27 '14 at 15:43
  • I'm not sure what the problem is. If you have no problem seeing that $C_n(A)$ is a subset of $C_n(X)$, then it should be no problem to see that i) the empty chain is a chain whose trace lies in $A$, and ii) the difference of two chains whose trace lies in $A$ is a chain whose trace lies in $A$. If you have a problem with $C_n(A)\subset C_n(X)$, it may be right for you to consider the map on the chain groups induced by the inclusion $\iota \colon A \hookrightarrow X$, which gives embeddings $\iota_\ast \colon C_n(A) \hookrightarrow C_n(X)$ and view $C_n(X)/C_n(A)$ as a slightly incorrect ... – Daniel Fischer Mar 27 '14 at 15:50
  • abbreviation for $C_n(X)/\iota_\ast(C_n(A))$. Note however, that with $C_n(A) = { c\in C_n(X) : \operatorname{Tr} c \subset A}$, we have a true subset and hence subgroup relation. The induced homomorphism in 2) is indeed the one from the factor theorem, if I have abelian groups $G,H$ and a homomorphism $\varphi\colon G\to H$, then for every subgroup $N \subset \ker\varphi$ I have an induced homomorphism $G/N \to H$ (for non-abelian groups, one must also require that $N$ be a normal subgroup explicitly). – Daniel Fischer Mar 27 '14 at 15:54
  • Ok I'm sorry that I didn't really elaborate on my problem. I understand that $C_n(A) \subset C_n(X)$. What's this trace all about? – ireallydonknow Mar 27 '14 at 15:59
  • Trace, image, whatever you call it, the trace of a chain is the union of the images of all singular simplices occurring with a nonzero coefficient in the chain. Okay, if you have no problem with $C_n(A) \subset C_n(X)$, are there any problems with i) $0 \in C_n(A)$, ii) $c_1,c_2 \in C_n(A) \implies c_1+c_2\in C_n(A)$, iii) $c\in C_n(A)\implies -c\in C_n(A)$? If so, which? – Daniel Fischer Mar 27 '14 at 16:09
  • Hi, sorry for the late reply. Ah, that's where I got lost. (i) onwards, I don't really understand. – ireallydonknow Mar 28 '14 at 13:27
  • $C_n(A)$ is the subset of $C_n(X)$ where the coefficients of every singular simplex whose image is not entirely contained in $A$ are $0$. The $0$ chain is the chain where the coefficient of every singular simplex is $0$, in particular the coefficient of the simplices not entirely contained in $A$ are $0$, hence $0 \in C_n(A)$. If the coefficient of the singular simplex $\sigma$ is $0$ in the two chains $c_1$ and $c_2$, then it is also $0$ in $c_1+c_2$ and in $-c_1$ - the coefficient is the sum of the coefficients in $c_1$ and $c_2$ resp. the negative of the coefficient in $c_1$. Since that – Daniel Fischer Mar 28 '14 at 13:49
  • holds for every singular simplex whose image isn't entirely contained in $A$, it follows that all singular simplices whose coefficient in $c_1 + c_2$ resp. $-c_1$ is nonzero, must have their image entirely in $A$, and thus $c_1 + c_2 \in C_n(A)$ and $-c_1\in C_n(A)$. – Daniel Fischer Mar 28 '14 at 13:52