If $(X, A)$ is a Good Pair then $i_*:H_n(X, A)\to H_n(X, V)$ is an Isomorphism 
Definition. Let $A$ be a closed subspace of a topological space $X$. We say that $(X, A)$ is a good pair if there is a neighborhood $V$ of $A$ in $X$ which deformation retracts to $A$.

In the proof of Proposition 2.22 in Hatcher's Algebraic Topology, the author shows the following:

Fact. Let $(X, A)$ be a good pair and $V$ be a neighborhood of $A$ in $X$ which deformation retracts to $A$. Then the inclusion map $i:(X, A)\to (X, V)$ induces an isomorphism $i_*:H_n(X, A)\to H_n(X, V)$ for all $n$.

The proof given in the book is a one liner. We just use the long exact sequence for the triple $(X, V, A)$ along with the fact that $H_n(V, A)$ is $0$ for all $n$.
But I have no intuition for the above solution and as of now the use of the long exact sequence seems like an algebraic trick to me. So I tried a more direct approach.

Claim. $i_*$ is injective.

Proof. Let $\sigma\in C_n(X)$ be arbotrary with $\partial \sigma\in C_{n-1}(A)$ and suppose that $i_*([\bar \sigma])=0$ (We use $\bar \sigma$ to denote the coset of $\sigma$ in $C_n(X, A)$ and $[\bar \sigma]$ to denote the homology class of $\bar \sigma$). So $\bar \sigma$ is a boundary in $Z_n(X, V)$. This means that there is $\tau\in C_{n+1}(X)$ such that $\bar \sigma=\partial\bar \tau$ in $C_n(X, V)$ (the meaning of the 'bars' have been suitably changed). So $\sigma -\partial \tau\in C_n(V)$. Note that $\partial (\sigma -\partial \tau)\in C_{n-1}(A)$, which means that $\overline{\sigma-\partial \tau}\in Z_n(V, A)$. But since $H_n(V, A)=0$, we have $Z_n(V, A)=B_n(V, A)$. Thus $\overline{\sigma-\partial \tau}=\partial \bar\theta$ for some $\theta\in C_{n+1}(V)$. This leads to $\sigma-\partial(\tau+\theta)\in C_n(A)$, giving $\bar \sigma=\partial\overline{(\tau+\theta)}$ in $C_n(X, A)$. This shows that $[\bar \sigma]=0$ in $H_n(X, A)$ and we are done.

The Problem. I am having trouble showing the surjectivity of $i_*$.

Can somebody please help me with this and also, if possible, throw some light on the use of the long exact sequence.
Thanks.
 A: The proof of the surjectivity follows the same idea. 
Let $\sigma + \text{Im } \partial$ be an element of $H_n(X,V)$. Since $\sigma \in Z_n(X,V)$, we have $\partial \sigma \in C_{n-1}(V)$. Abusing the notation, we can consider $\partial \sigma$ as an element of $C_{n-1}(V,A)$. Note that $\partial \partial \sigma = 0$, in particular $\partial \partial \sigma \in C_{n-2}(A)$. Hence $\partial \sigma \in Z_{n-1}(V,A)$. 
Since $H_{n-1}(V,A)=0$, we must have $\partial \sigma \in B_{n-1}(V,A)$. That is, there exists $\tau \in C_n(V)$ such that $\partial \sigma - \partial \tau = z \in C_{n-1}(A)$.
Now $(\sigma - \tau) + \text{Im } \partial = \sigma + \text{Im } \partial $ in $H_n(X,V)$ and $\partial(\sigma - \tau) \in C_{n-1}(A)$. Therefore $(\sigma - \tau) \in Z_n(X,A)$ and so it makes sense to consider $(\sigma - \tau) + \text{Im } \partial$ as an element of $H_n(X,A)$. And clearly 
$$ i_*((\sigma - \tau) + \text{Im } \partial) = (\sigma - \tau) + \text{Im } \partial = \sigma + \text{Im } \partial $$

The use of the long exact sequence is not an algebraic trick, it is a way to summarize conveniently what we did (and more!). Note that to show injectivity you used that $H_n(V,A)=0$ and to show surjectivity I used $H_{n-1}(V,A)=0$. If you look at the proof (or even better, do it) that the sequence
$$ \ldots \to H_n(V,A) \stackrel{f}{\to} H_n(X,A) \stackrel{i_*}{\to} H_n(X,V) \stackrel{g}{\to} H_{n-1}(V,A) \to \ldots $$
is exact, showing that the kernel of $i_*$ equals the image of $f$ is practically the same idea you used for the injectivity of $i_*$. What you did is a particular case. Same thing for the image of $i_*$ and the kernel of $g$. 
The proof of the existence of this long exact sequence is often shown via a very general argument that a short exact sequence of complexes induces a long exact sequence of homology groups. But if you unpack the proof, it comes back to a generalization of our arguments.
It pays to get used to long exact sequences and how to use them. Arguments are often shorter with them because they already contain a lot of information in their exactness. I hope this helps.
