Show that the abelianization functor is right exact I have a homework question that I do not understand for an abstract algebra class.  We have covered the first couple chapters in Dummit and Foote (on groups).  Also, I know nothing of functors besides what is given below.  We were told that we shouldn't need the formal definition of a functor.

A functor $F$ is right exact if, given a sequence $A \rightarrow B \rightarrow C \rightarrow 0$ that is exact at $B$ and $C$, $F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow 0$ is exact at $B$ and $C$.
Show that the abelianization functor F that maps $A \rightarrow B$ to $A/[A,A] \rightarrow B/[B,B]$ is right exact.

I guess my question is really, what the heck is this saying!?  This notation is a bit new, so right now I'm interpreting this as $A$, $B$, and $C$ being groups with each arrow representing a homomorphism.  If this is the case, what exactly is $F(A)$?  It looks like $F$ maps a homomorphism to another homomorphism, so what is $F(A)$ if A is a group?
Any help clarifying this question would be greatly appreciated.
 A: Given a homomorphism of groups $f:G\to H$ one can compose with the projection map $H\to H^{\text{ab}}$ to get a homomorphism $G\to H^{\text{ab}}$. But, since $H^{\text{ab}}$ this map factors through $G^\text{ab}$ giving us a map $f^{\text{ab}}:G^\text{ab}\to H^\text{ab}$. 
Explicitly
$$f^\text{ab}(a+[G,G])=f(a)+[H,H]$$
You need to show the following:
If 
$$A\xrightarrow{f} B\xrightarrow{g} C\to 0$$
is exact, then
$$A^\text{ab}\xrightarrow{f^\text{ab}}B^\text{ab}\xrightarrow{g^\text{ab}}C^\text{ab}\to 0$$
is also exact.
I leave the actual verification of this to you. It's not so bad, but feel free to let me know if you'd like further hints, which I will be glad to supply.
A: This was for a comment for the questions and comments by Steve years ago under Alex's Answer. However, since it may be a bit long and directly related to the question, I think it may worth making an answer on it own right.
I would first quote the OP's work here,

$f^{ab}(A^{ab}) = f^{ab}(A + [A, A]) = f(A) + [B,B] = \text{ker}(g)+[B,B] = \text{ker}(g)/[B,B] = \text{ker}(g^{ab})$.

In fact, the notation $\text{ker}(g)/[B,B]$ doesn't make much sense to me, since $\text{ker}(g)$ does not necessarily contain the entire $[B, B]$, so I guess what it really means is $(\text{ker}(g) + [B, B])/[B, B]$, but this is really just $\text{ker}(g) + [B, B]$ in $B^{ab}$ as a subgroup. $\text{ker}(g)+[B,B] = \text{ker}(g^{ab})$ is in fact correct, but here the surjectivity of $g$ is needed, which by exactness of the sequence $A\xrightarrow{f}B\xrightarrow{g}C\rightarrow 0$ is true.
Claim that, for the exact sequence
\begin{equation}
A\xrightarrow{f}B\xrightarrow{g}C,
\end{equation}
in general, $\text{ker}(g)+[B,B] \subsetneq \text{ker}(g^{ab})$. In fact, by definition,
\begin{equation}
b + [B, B]\in \text{ker}(g^{ab})\Leftrightarrow g^{ab}(b + [B, B]) = 0 \Leftrightarrow g(b) + [C, C] = 0\text{ 
 in  }C^{ab} \Leftrightarrow g(b)\in [C, C].
\end{equation}
On the other hand,
\begin{equation}
b + [B, B]\in \text{ker}(g)+[B,B] \Leftrightarrow \exists\ b'\in[B, B], g(b + b') = 0.
\end{equation}
Now it boils down to compare $g([B, B])$ and $[C, C]$. It is straightforward that $g([B, B]) \subseteq [C, C]$. As for the other direction, it is not necessarily true if we don't assume $g$ to be surjective, which explains your example $0\rightarrow S_2\xrightarrow{h}D_8$. If we do have $g$ being surjective, as implied by the original exact sequence $A\xrightarrow{f}B\xrightarrow{g}C\rightarrow 0$, $g([B, B])$ contains all commutators of C. Moreover, since $g$ is surjective and $[B, B]$ is normal, $g([B, B])$ is normal, so it contains $[C, C]$. Now we have established $g([B, B]) = [C, C]$, and then it follows $\text{ker}(g)+[B,B] = \text{ker}(g^{ab})$, which completes the proof.
