Seeking an intuitive explanation of the Mapping Class Group For a surface $S$ the mapping class group $MCG(S)$ of $S$ is defined as the group of isotopy classes of orientation preserving diffeomorphisms of $S$:
$$MCG(S)=Diff^+(S)/Diff_0(S).$$
I understand this definition as well as all of its component pieces. What I don't understand is why this quotient is a natural thing to study.
Specifically, I can see why the full diffeomorphism group $Diff(S)$ would be natural to study, and if $S$ happens to be orientable, I can see why it would be reasonable to restrict ones attention to $Diff^+(S)$. However, I don't see why the quotient is a natural or intuitive next step. Is there a good explanation why diffeomorphisms that are isotopic to the identity are 'uninteresting'? Thanks!

To explain why this is interesting, let's look at an analysis of a particular case, namely $S = T^2$. 
From algebraic topology we learn about the homology functor, from which we deduce that there is a homomorphism
$$\text{Diff}^+(T^2) \to \text{Aut}(H_1(T^2;\mathbb{Z})) \approx \text{SL}(2,\mathbb{Z})
$$
Thinking about the construction of $T^2$ as the quotient $\mathbb{R}^2 / \mathbb{Z}^2$, and thinking about the fact that elements of the group $\text{SL}(2,\mathbb{Z})$ act on $\mathbb{R}^2$ preserving the orbits of the additive action of $\mathbb{Z}^2$, one quickly concludes that the above homomorphism is surjective.
Once you've gone that far, nothing is more natural from an algebraic perspective to ask: what is the kernel of that homomorphism? Because, once you figure out that the kernel is $\text{Diff}_0(T^2)$, you get an amazing theorem:
$$\text{MCG}(T^2) \approx \text{Diff}^+(T^2) / \text{Diff}_0(T^2) \approx \text{SL}(2,\mathbb{Z})
$$
And once you have that theorem in front of you, what is more natural than to generalize, and to ask "What can I say about $\text{Diff}^+(S)/\text{Diff}_0(S)$ in the general case?"
