# Limit of sequence of abelian groups

I'm in Awodey's Category Theory, page 118, exercise 13.

Consider sequences of abelian groups (in the category of monoids) $$M_0 \to M_1 \to M_2 \dots$$ $$N_0 \gets N_1 \gets N_2 \dots$$

Determine whether the four limits/colimits of $M_n, N_n$ are abelian groups.

I can do them for the $M_n$ sequence, I think: the limit is (isomorphic to) $M_0$ and hence is an abelian group. The colimit $C$ is an abelian group, because for any element $x$ in the colimit, we can find $n$ and $\alpha$ such that $i_n(\alpha) = x$, where $i_n$ is the inclusion $M_n \to C$. Therefore we can view $x$ as being in all sufficiently large-$n$ $M_n$; so any pair of elements may be viewed in some $M_n$, which is an abelian group. Therefore all the abelian group properties must hold.

(Sadly I think I can't say by duality that the $N_n$ sequence has the same properties, because the opposite category of Mon is not Mon.)

However, similar reasoning to the $M_i$ case won't work directly on the $N_i$ case, because we could always end a sequence with $N_0 = \{ e \}$. Therefore we can't view an element of the limit or colimit as being in "sufficiently large $N_i$" and do anything non-trivial with that information. It feels like the colimit should be (isomorphic to) $N_0$ and hence abelian, but how do I approach the limit?

The colimit of the $N_i$ is indeed $N_0$ (together with the given morphism $N_i\to N_0$), for if $X$ is a monoid together with homomorphisms $f_i\colon N_i\to X$ with $f_i\circ d=f_{i+1}$ then of course this factors nicely over $N_0$.
Let $N$, together with homomorphisms $\pi_i\colon N\to N_i$ be the limit. Mapping $x\mapsto -x$ in each $N_i$ induces an involutory automorphism $x\mapsto \bar x$ of $N$. As all $x\bar x$ map to $0$ in all $N_i$, we must have $N=N/\langle x\bar x\rangle$ so that - as expected - $\bar x$ is inverse to $x$ in $N$ and $N$ is a group. Since the map $x\mapsto x^{-1}=\bar x$ is still an automorphism of $N$, $N$ must be abelian.