Are there any non-obvious colimits of finite abelian groups?

Solution 1:

I claim that the colimit of the sequence of inclusions $\frac{1}{1!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \frac{1}{2!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \frac{1}{3!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \dotsc$ exists in $\mathsf{FinAb}$. Namely, it is zero, in contrast to the colimit in $\mathsf{Ab}$, which is $\mathbb{Q}/\mathbb{Z}$. This is equivalent to the following statement:

Let $A$ be a finite abelian group containing a sequence of elements $x_1,x_2,\dotsc$ such that $n! x_n = 0$ and $(n+1) x_{n+1} = x_n$ for all $n \geq 1$. Then $x_n=0$ for all $n \geq 1$.

Proof. Because of $x_{n+1}=0 \Rightarrow x_n=0$, it suffices to prove $x_n=0$ for infinitely many $n$. So we may assume that $\mathrm{ord}(A)$ divides $n+1$. But then $x_n=(n+1) x_{n+1} = 0$ by Lagrange's Theorem. $\square$

It follows that the forgetful functor $U : \mathsf{FinAb} \to \mathsf{Ab}$ preserves coproducts and coequalizers, but it does not preserve colimits of sequences, and hence it is not cocontinuous.

(The statement that a functor, which preserves coproducts and coequalizers, is cocontinuous, can be found in many books and online references. These should be corrected.)

PS: Sorry for answering my own question.