Let \(A,B\) be \(R\)-modules. Show that \[ \Hom_R(A,B) \] is an abelian group under pointwise addition \[ (f+g)(a)=f(a)+g(a). \]
Solution
If \(f,g\in\Hom_R(A,B)\), then \(f+g\) is \(R\)-linear because \[ (f+g)(a+a')=f(a)+f(a')+g(a)+g(a') =(f+g)(a)+(f+g)(a') \] and \[ (f+g)(ra)=rf(a)+rg(a)=r(f+g)(a). \] Thus pointwise addition is closed.
Associativity and commutativity follow pointwise from the abelian-group law on \(B\): \[ (f+(g+h))(a)=f(a)+g(a)+h(a)=((f+g)+h)(a), \] \[ (f+g)(a)=f(a)+g(a)=g(a)+f(a)=(g+f)(a). \]
The zero map is the identity element, and the inverse of \(f\) is the map \[ (-f)(a)=-f(a), \] which is again \(R\)-linear.
Therefore \[ (\Hom_R(A,B),+) \] is an abelian group.