The image of the boundary homomorphism in the long exact sequence of homology groups coincides with the image of Hurewicz map. I have trouble understanding Hatcher's claim that the image of the boundary homomorphism in the long exact sequence of homology groups coincides with the image of Hurewicz map at the top of this page(Ex.4.2.23, where Y is obtained from X by attaching cells of dimension $n+2$). 
It doesn't seem follow directly from the definition of Hurewicz map and/or definition of the boundary map. I think we need to find an appropriate interpretation of the boundary map(which can be proved to be equivalent to the definition of it).
Updated on 4/11:
Let $\alpha$ be a fixed generator of $H_n(D^{n+1}, \partial D^{n+1})\cong \mathbb Z$. Then the image of the Hurewicz map is $\{f_*(\alpha):[f]\in \pi_{n+1}(X) \}$ where $f_*$ is induced by $f:(D^{n+1}, \partial D^{n+1})\to (X,x_0)$.
On the other hand, the boundary map $\partial:H_{n+2}(Y,X) \to H_{n+1}(X)$ is defined by the diagram chasing. I can't see the connection between them from the definitions. If you would like to interpret $\partial$ in another way, please explain why this new interpretation coincides with the definition of $\partial$ defined by the diagram chasing.   
 A: First, the topological point of view:
Let $[\sigma] \in H_n(X,A)$. This is some chain in $X$ with boundary inside of $A$.But then it must be a cycle, but it need not bound anything entirely in $A$, so it could be a nonzero representative in $H_{n-1}(A)$. To say the exact same thing in a little more detail:  with singular homology: map $\sigma \mapsto X$ so that $\partial  \sigma$ (restriction to the boundary of $S^n$) maps entirely into $A$. This boundary becomes an element of $H_{n-1}(A)$.
Now, what was the algebraic point of view? 
I'm going to call all the boundary maps $\partial$, I hope this is not too confusing, I'll reserve $\partial_*$ for our big question. Take
$$0 \to C(A) \to C(X) \to C(X,A) \to 0$$
as a short exact sequence of chains.
Pick some guy $\alpha$ in the kernel of $\partial:C_n(X,A) \to C_{n-1}(X,A)$. This means that the boundary of $\alpha$ was some chain regarded as zero if its boundary was  in $A$. 
We follow the snake: But then $\alpha= j_*(\alpha^{\prime})$, which was just a projection, so we consider $\alpha^{\prime}$ to be that chain whose boundary was totally in $A$. But then, $\partial \alpha^{\prime} \in \ker j_*(C_{n-1}(X))$. Algebraically, we use exactness here to pullback to some element in $C_{n-1}(A)$ which is unique by injectivity, but the geometric picture is clear, we just consider that cycle in $A$, call it $\alpha^{\prime \prime}. We use this element to take homology.
So, $\partial_*(\alpha)=\alpha^{\prime \prime}+\mathrm{Im}\partial_n$.
I hope this stitched the two together, whether by diagram chasing, or by geometric considerations, the map is quite natural (I actually find this to be quite miraculous.)
