3
$\begingroup$

For any natural number $n$, let $i_{n},j_{n}:X_{n}\rightarrow X_{n+1}$ be a pair of monomorphisms of simplcial sets.

Define $$X=\operatorname*{colim}_n \{\cdots X_n \rightarrow_{i_n} X_{n+1}\cdots \} $$ and $$Y=\operatorname*{colim}_n \{\cdots X_n \rightarrow_{j_n} X_{n+1}\cdots \}$$

Question: I'm looking for an example where $X$ is not isomorphic to $Y$ as a simplicial set.

$\endgroup$
3
  • $\begingroup$ It would be enough to find an example in sets, categories or groups, cause the latter embeds into simplicial sets (preserving directed colimits and monomorphisms). In sets, unluckily, there are no examples, because the two infinite unions always have the same cardinality. $\endgroup$ Jun 22, 2019 at 9:10
  • 5
    $\begingroup$ @Andrea, why don't you write that up? You could take for example $\mathbb{Z} \to \mathbb{Z} \to \ldots$ where in one case it's all identity maps, and in the other it's all multiplication by 2 (in groups). $\endgroup$
    – Todd Trimble
    Jun 22, 2019 at 11:29
  • $\begingroup$ Yeah, sure! I thought about that in ordinals with multiplication by 2 in omega, but didn't realize we could do it in groups more easily (in ordinals, honestly, I wouldnt know how to show it)! $\endgroup$ Jun 23, 2019 at 0:19

1 Answer 1

2
$\begingroup$

Collecting the comments, we have the following two observations:

  1. Consider the functor $NB: Grp \to Cat \to sSet$, where the first functor,B, takes a group to the category with one object and endomorphisms equal to the group, and the second,N, is the nerve. Note that:

A) B factors through Cat_1, the categories with one object. We have a right adjoint to B given by the automorphism group of the only object involved. Thus $B$ is a left adjoint composed with a full inclusion $Cat_1 \to Cat$. The latter evidently preserves colimits. Moreover, B takes injectives to maps that are injective on both objects and arrows.

B) The nerve is given by the formula $NC_n =Hom([n], C)$. Note that the objects $[n]$ are finite, so that morphisms from $[n]$ to a directed colimit $\{X_k\}$ indeed end up on one of the objects $X_k$. A rephrasing of this gives exactly that N commutes with directed colimits. Also, it is evident from the formula that N takes injective on both objects and arrows to injective levelwise, which are the mono of sSets.

On balance, we can reduce to find an example in Groups.

  1. Take, as wisely suggested by Todd Trimble, the two sequences $$ \mathbb{Z} = ... = \mathbb{Z} = ... $$ $$ \mathbb{Z} \overset{2*}{\to} ... \mathbb{Z} \overset{2*}{\to}... $$

The first sequence evidently has colimit Z. The other one is not: indeed, we show that there are no non-zero morphisms from the second colimit, W, to Z modulo 2. Suppose there is one collection $\{f_k\}$ with one of them non zero. Say $f_j(x)=1$ for some j,x. Then $$0= 2f_{j+1}(x) = f_{j+1}(2x) = f_j(x) = 1$$ A contradiction.

$\endgroup$
2
  • 2
    $\begingroup$ Thanks! The colimit in the second case is also known as a Prüfer group: en.wikipedia.org/wiki/Pr%C3%BCfer_group. The only thing I'd add to your proof is that the inclusion $Grp \to sSet$ is fully faithful, so that if the colimits on the simplicial set side were isomorphic, they'd have to be isomorphic on the group side, leading to a contradiction as you explained. $\endgroup$
    – Todd Trimble
    Jun 23, 2019 at 1:42
  • $\begingroup$ Yes! You are right. Both are fully faithful! $\endgroup$ Jun 23, 2019 at 7:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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