7

It seems to me that that this is true:

In the category of Sets, transfinite composition of monomorphisms is again a monomorphism.

Explicitly, given a $\lambda$-sequence $$X_0 \xrightarrow{f_1} X_1 \cdots $$ Each map $f_i$ is a monomorphism, then is the canonical $$ g:X_0 \rightarrow colim_{i < \lambda} X_i$$ A monomorphism?

Is this true more generally in other categories?

My thoughts: In sets there are explicit description of colimit. So I suppose we do this by some kind of ordinal induction?

We know this is true for successor ordinals. $\beta \le \lambda$. If $\beta\le \lambda$ is a limit ordinal, then we have $$g:X_0 \rightarrow X_\beta \simeq colim_{j< \beta} X_j$$ By property of ordinals, being totally ordered, exists some $k <\beta$, such that $$f_k(x)=f_k(y) \in X_k$$ By hypothesis, $x=y$.

Community
  • 1
Bryan Shih
  • 9,934
  • Since I don't know these things instinctively, take $X_n=\Bbb N\setminus{0,\dots,n-1}$, and $f_n$ given by $f_n(i)=i+1$. What would be the colimit of this, just over $\omega$? – Asaf Karagila Aug 03 '19 at 02:04
  • @AsafKaragila: The colimit is just $\mathbb{N}$: up to isomorphism, all the $X_n$'s are just $\mathbb{N}$ with the maps being the identity map. – Eric Wofsey Aug 03 '19 at 02:04
  • @Eric: Oh. Then replace $X_n$ by $\Bbb N$ itself, and just became "less and less surjective"? – Asaf Karagila Aug 03 '19 at 02:05
  • @AsafKaragila: In that case the colimit would be $\mathbb{Z}$: you can identify each $X_n$ with the integers greater than $-n$ and the maps then just become inclusions. – Eric Wofsey Aug 03 '19 at 02:07
  • More generally, the way you compute a directed colimit of injections of sets is to replace it with an isomorphic diagram where the maps are inclusions, and then take the union. – Eric Wofsey Aug 03 '19 at 02:07
  • @Eric: Ah, I see! Thanks! – Asaf Karagila Aug 03 '19 at 02:09
  • @Eric, so your claim perhaps prove the problem in greater generality? I have noticed another confusion - how does one guarantee that there is a unique compoisite $X_1 \rightarrow X_\beta$, for any ordinal $\beta< \lambda$? I suppose this where the condition $colim_{j<\alpha} X_j \simeq X_\alpha$ for $\alpha$ a limit ordinal comes in. – Bryan Shih Aug 03 '19 at 02:12
  • I think there is a problem here for infinite $\lambda$. For instance, say you have $\lambda=\omega+1$. In order to define the colimit, we need to have maps $X_i\to X_{\omega}$. for each $i\in\omega$. How are these defined? The maps $X_i\to X_j$ are clear when there is no limit ordinal between $i$ and $j$, but when there is one, it is not clear what the definition is yet. – Arturo Magidin Aug 03 '19 at 02:14
  • @ArturoMagidin: The link provided in the question gives the full definition. – Eric Wofsey Aug 03 '19 at 02:16
  • @EricWofsey: So, at each limit ordinal you just take the colimit of the previous ones and use that? Okay. – Arturo Magidin Aug 03 '19 at 02:22
  • 2
    @ArturoMagidin: Right, each limit object must be the colimit of the previous ones, and so the maps into it are just the inclusions into the colimit. – Eric Wofsey Aug 03 '19 at 02:25

1 Answers1

12

Yes, this is true for sets, and your proof sketch is basically correct. By induction, you show the transfinite composition $X_0\to X_i$ is injective for all $i$. Successor steps are trivial (you just compose with the injection $X_i\to X_{i+1}$). At limit steps, you use the fact that if two elements of $X_i=\operatorname{colim}_{j<i} X_j$ are equal, they must be represented by equal elements of $X_j$ for some $j<i$. So if two elements of $X_0$ map to the same element of $X_i$, they must have mapped to the same element of $X_j$ for some $j<i$, but by the inductive hypothesis the map $X_0\to X_j$ is injective.

It is not true in arbitrary categories. For instance, in the category of finite abelian groups, let $X_n=\mathbb{Z}/(p^n)$ for some $p>1$ with maps $X_n\to X_{n+1}$ given by multiplication by $p$. All these maps are injections, but the colimit $X_\omega$ is trivial, since there are no nontrivial maps from the diagram to any finite abelian group (an element in the image of such a map would have to be annihilated by some power of $p$ but also infinitely divisible by $p$, which is not possible for a nonzero element in a finite abelian group). So in particular, the transfinite composition $X_1\to X_\omega$ is not monic.

Eric Wofsey
  • 342,377