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On page 115 of Hatcher, he claims that a relative cycle $\alpha$ is trivial in $H_{n}(X,A)$ iff it is a relative boundary: $\alpha=\partial \beta + \gamma$ for some $\beta \in C_{n+1}(X)$ and $\gamma \in C_{n}(A)$.

He then says that these properties make precise the intuitive idea that $H_{n}(X,A)$ is homology of $X$ modulo $A$.I've been trying to prove the highlighted statement to gain some intuition about this, but I cannot figure out how to prove it, nor do I really have intuition about why this makes precise the idea that it is homology of $X$ mod $A$. Can someone please help me prove this and possibly explain the inutition?

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    Doesnt it follow directly from the definition of $C_n(X,A)$ as a quotient? – 2017-02-04

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Okay, so first off, how do we define $H_n(X,A)$? Well, we start with relative chain groups $C_n(X,A):=C_n(X)/C_n(A)$, and we note that our normal boundary map descend into a well-defined map on the relative chain groups:

$$\cdots\longrightarrow C_{n+1}(X,A)\overset{\partial}{\longrightarrow} C_n(X,A)\overset{\partial}{\longrightarrow} C_{n-1}(X,A)\longrightarrow\cdots$$

Then we take the homology of this complex to get $H_n(X,A)$.

Okay, so now let $\alpha\in C_n(X)$. First let's suppose the image of $\alpha$ is trivial in $H_n(X,A)$. We must ask ourselves, how can things be trivial in $H_n(X,A)$? Well, first they can be trivial in $C_n(X,A)$, which means $\alpha\in C_n(A)$, or they can be in the image of $\partial$, i.e. $\alpha=\partial(\beta)$ for some $\beta\in C_{n+1}(X)$. Also, any sum of elements that are trivial in one of these two ways will be trivial in $H_n(X,A)$ as well. And by the way we've defined things, these are actually the only possibilities. Therefore if $\alpha$ is trivial we can write it as $\alpha=\partial(\beta)+\gamma$ for $\beta\in C_{n+1}(X)$ and $\gamma\in C_n(A)$.

Now, on the other hand, if $\alpha=\partial(\beta)+\gamma$ in the first place, then by definition the image in $C_n(X,A)$ is equal to the image of $\partial(\beta)$ (because $\gamma$ is now trivial), and the further image in $H_n(X,A)$ is trivial because the image of $\partial(\beta)$ is trivial in view of homology.

As for intuition, what Hatcher is trying to say by "homology modulo $A$", is he's just basically saying we look at the normal definition of homology, i.e. boundaries are trivial, plus we add this condition that anything occurring in $A$ is trivial as well. Hence, we are going "modulo $A$".