1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Just to be completely clear, then: all four of the colimits and limits specified in the question are in fact abelian groups? (I'll read this over again in the morning, and will probably accept it then.) $\endgroup$ – Patrick Stevens Sep 27 '15 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.