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 $(-s)(a)(-s)(b)=(-s(a))(-s(b))=s(a)s(b)$.
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.