I know that colimit preserves epimorphisms.

Consider the special case where

  1. The diagrams are indexed by a directed set $I$,
  2. We are in the category of certain algebraic structures, such as $\mathbf{Ab}, \mathbf{Ring}, \mathbf{Set}$, $\mathbf{Mod_R}$, where colimit can be explicitly constructed as the disjoint union quotient an equivalence.
  3. The category we are considering has free objects, i.e. the forgetful functor to $\mathbf{Set}$ has a left adjoint. In this case monomorphisms are exactly the injective maps.

In this case, it seems to me that colimit also preserves monomorphisms. Given monomorphisms $f_i: A_i\to B_i$, we show that $f: \varinjlim A\to \varinjlim B$ is a monomorphism.

If two elements $[(i,a)],[(j,b)]$ are mapped to the same class, we can find $i\le k, j\le k$ such that $[(k,a')] = [(i,a)]$ and $[(k,b')] = [(j,b)]$ where $[(k,a')]$ and $[(k,b')]$ are mapped to the same class $[(k,f_k(a') = f_k(b')]$. But since $f_k$ is injective, we must have $a' = b'$, and that $[(i,a)] = [(j,b)]$.

Is it correct? The motivation of the question comes from an exercise where I try to prove that a morphism of sheafs $\Phi:\mathscr{F}\to\mathscr{G}$ is injective on open sets $\Phi_U:\mathscr{F}(U)\to \mathscr{G}(U)$ implies that it is injective on stalks.

  • 1
    $\begingroup$ Co-limit preserves exactness in $\mathbf{Mod_R},$ where $R$ is a commutative ring with identity. See math.stackexchange.com/questions/121122/… $\endgroup$
    – Krish
    Jan 23 '15 at 6:58
  • $\begingroup$ @Krish That's true, but I'm in particular interested in $\mathbf{Set}$. $\endgroup$
    – mez
    Jan 23 '15 at 7:26
  • $\begingroup$ What is the map $f$ that you state is injective? $\endgroup$
    – tracing
    Jan 25 '15 at 20:16
  • $\begingroup$ @tracing It is $f: \varinjlim A_i\to \varinjlim B_i$ induced by $f_i:A_i\to B_i$ $\endgroup$
    – mez
    Jan 25 '15 at 22:39

Good observation!

Indeed it happens often that filtered colimits preserve finite limits, and as for a morphism $f: X\to Y$ being mono is equivalent to $id, id: X\rightrightarrows X$ being a pullback of $(f,f)$, in these situations monomorphisms are preserved.

You find this imposed as a condition in topos theory, where a geometric morphism $f: {\mathcal E}\to{\mathcal F}$ consists of an adjunction $f^{\ast}: {\mathcal E}\rightleftarrows {\mathcal F}: f_{\ast}$ in which additionally $f^{\ast}$ is required to preserve finite limits.

See http://ncatlab.org/nlab/show/geometric+morphism

As you already observed, taking ${\mathcal F}=\text{Sh}(X)$, ${\mathcal E}=\text{Sh}(\text{pt})$ and $f: \text{Sh}(\text{pt})\to\text{Sh}(X)$ coming from a point $x\in X$, then the extra condition on $f^{\ast} = (-)_x$ means that the formation of stalks preserves finite limits, which you already observed in case of monomorphisms.

  • $\begingroup$ $f$ is a mono if and only if $\mathrm{id}_X,\mathrm{id}_X \colon X \rightrightarrows X$ is the pullback of $f,f \colon X \rightrightarrows Y$. This is different of the condition you give (if I'm not mistaking, your condition is vacuous). $\endgroup$
    – Pece
    Jan 23 '15 at 9:41
  • $\begingroup$ Omg... Sorry for that - and thank you, Pece! $\endgroup$
    – Hanno
    Jan 23 '15 at 9:46

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