$t-s$ is a group homomorphism? 
For each pair of group homomorphisms $s,t:A\to B$ in $\mathbf{Ab}$, there is a homomorphism $t-s:A\to B$.

I've been trying to understand how to define this map to make it a group homomorphism. One obvious way is $(t-s)(a)=t(a)-s(a)$. But it doesn't seem to be a homomorphism:


*

*$(t-s)(ab)=t(ab)-s(ab)=t(a)t(b)-s(a)s(b)$

*$(t-s)(a)\cdot(t-s)(b)=(t(a)-s(a))(t(b)-s(b))\\=t(a)t(b)-s(a)t(b)-t(a)s(b)+s(a)s(b)$
Are these two elements equal? This is so iff $t(a)s(b)=-s(a)t(b)$, which is I don't know why should hold.
And even more narrowly, is $-s$ a homomorphism? I would think a reasonable definition is $(-s)(a)=-s(a)$. But again $(-s)(ab)=-s(ab)=-s(a)s(b)$ whereas $(-s)(a)(-s)(b)=(-s(a))(-s(b))=s(a)s(b)$.
 A: You can use multiplicative notation, but then you should be consistent. And as you will see below, it just begs you to use additive notation instead.
If $A$ and $B$ are abelian groups written multiplicatively, and $f\colon A\to B$ is a group homomorphism, then you have a group homomorphism which should be called $f^{-1}$, but unfortunately that notation is taken, defined by $a\mapsto f(a)^{-1}$. Essentially, it is the composition of the map $f$ with the homomorphism ${}^{-1}\colon B\to B$ that maps each element to its inverse, and hence is a homomorphism since it is a composition of two homomorphisms. As the notation $f^{-1}$ would be confusing at best, let us denote it instead by $f'$. So, if $f\colon A\to B$ is a homomorphism of abelian groups, then $f'\colon A\to B$ is given by $f'(a) = f(a)^{-1}$, and is a homomorphism because $B$ is assumed to be abelian; this because the map $b\longmapsto b^{-1}$ is a homomorphism if and only if $B$ is abelian.
Indeed, note that if $B$ is abelian, then
$$(bc)^{-1} = c^{-1}b^{-1} = b^{-1}c^{-1}$$
and conversely, if the map is a homomorphism, then for every $b,c\in B$, we have
$$bc= ((bc)^{-1})^{-1} = (b^{-1}c^{-1})^{-1} = (c^{-1})^{-1}(b^{-1})^{-1} = cb,$$
so the group is abelian.
Clarification: We go from $(xy)^{-1}$ to $y^{-1}x^{-1}$ because this holds in any group; and we can go from $(xy)^{-1}$ to $x^{-1}y^{-1}$ because we are assuming that $b\longmapsto b^{-1}$ is a homomorphism, so the inverse of $xy$ is the inverse of $x$ times the inverse of $y$. 
Given $f,g\colon A\to B$, we define $fg\colon A\to B$ to be the "pointwise multiplication" of $f$ and $g$, given by $(fg)(a) = f(a)g(a)$. For arbitrary groups this is not in general a group homomorphism, but if $B$ is abelian then this is indeed a homomorphism: we have:
$$(fg)(xy) = f(xy)g(xy) = f(x)f(y)g(x)g(y) = f(x)g(x)f(y)g(y) = (fg)(x)(fg)(y),$$
where the third equality holds because $B$ is abelian.
In this notation, we would not have "$t-s$", but rather we would have $ts'$, where $t$ and $s$ are given homomorphisms, and $s'$ is as defined above. This is indeed a homomorphism, either because both $t$ and $s'$ are and then applying the paragraph above, or checking directly because:
$$\begin{align*}
(ts')(xy) &= t(xy)s'(xy) &&\text{by definition of }ts'\\
&= t(x)t(y)(s(xy))^{-1} &&\text{by definition of}s'\\
&= t(x)t(y)(s(x)s(y))^{-1} &&\text{because }s\text{ is a homomorphism}\\
&= t(x)t(y)s(x)^{-1}s(y)^{-1} &&\text{because }B\text{ is abelian}\\
&= t(x)s(x)^{-1} t(y)x(y)^{-1} &&\text{because }B\text{ is abelian}\\
&= t(x)s'(x)t(y)s'(y) &&\text{by definition of }s'\\
&= (ts')(x)(ts')(y).
\end{align*}$$
Thus, $ts'(xy) = ts'(x)ts'(y)$, proving this is indeed a homomorphism. 

If you use additive notation throughout, you get
$$\begin{align*}
(t-s)(x+y) &= t(x+y) - s(x+y)&&\text{by definition of }t-s\\
&= t(x)+t(y) - (s(x)+s(y)) &&\text{because both }t\text{ and }s\text{ are homomorphisms}\\
&= t(x)+t(y) - s(x)-s(y)\\
&= t(x)-s(x) + t(y)-s(y) &&\text{because }B\text{ is abelian}\\
&= (t-s)(x) + (t-s)(y) &&\text{by definition of }t-s,
\end{align*}$$
proving that $t-s$ is a group homomorphism. 
A: What's the multiplication you're using there? Remember that you are working in Ab, you should be looking for an abelian group. Yet you're throwing yourself off with multiplication. In the special case $A=B$ this can make sense with the multiplication being given by composition, but it doesn't make sense for arbitrary groups. You do have the right idea tho, we want our addition to be pointwise which means $(t-s)(a)=t(a)-s(a)$ which is something that does in fact make sense in our given category. Following this, we have $(t-s)(0)=t(0)-s(0)=0-0=0$, so we see that this does in fact give us a new homomorphism $(t-s):A\to B$.
You had the right idea, just be sure to keep track of your data types.
