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It is probably a trivial question. But I don't see the answer (and I didn't find anywhere).

Given a (complete and cocomplete) category X and an object A of X, we can define the "undercategory" A/X. See http://ncatlab.org/nlab/show/under+category

I have already noticed that the coproduct of a set {i_l: A to X} is the "natural injection" of A in the colimit of the obvious diagram defined by the set.

I'm trying to understand how the product in A/X looks like, in terms of colimits, limits, products or coproducts of X.

I appreciate any help. Thank you very much!

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  • $\begingroup$ Instinctive guess: pushout? $\endgroup$
    – Yemon Choi
    Commented Nov 27, 2011 at 8:43
  • $\begingroup$ If K={i_l: A to X} is a family of objects in A/X. The "natural injection" morphism A to colimit of K is the coproduct of the objects in K. In particular, if K is a family of only two objects, then the natural morphism A to Pushut (K) is the coproduct... $\endgroup$
    – Fernando
    Commented Nov 27, 2011 at 9:12

2 Answers 2

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For a family of objects $A \to B_i$ in $A/X$, their coproduct is usually called the pushout of the morphisms $A \to B_i$ in $C$. It represents the functor $X \to \mathrm{Set}$, which maps $P$ to the set of families of morphisms $B_i \to P$ which "coincide" on $A$. Pushouts in $X$ may be constructed via coproducts and coequalizers in $X$. The idea is to take $\coprod_i B_i$ and then identify $A \to B_i$ with $A \to B_j$.

The products in $A/X$ are more easy: The forgetful functor $A/X \to X$ preserves (also creates) them. This means that for a family of objects $f_i : A \to B_i$ the product is given by $f : A \to \prod_i B_i$, where $f$ is defined by $\mathrm{pr}_i f = f_i$.

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Here is another perspective: the undercategory $A \downarrow \mathcal{X}$ is the category of algebras (Eilenberg-Moore category) for the monad whose underlying functor takes an object $X$ to $A \sqcup X$. (The unit of the monad is the coproduct inclusion $X \hookrightarrow A \sqcup X$; the multiplication is $\nabla_A \sqcup X: A \sqcup A \sqcup X \to A \sqcup X$.) Then we can just quote the result that a monadic functor

$$Alg_M \to \mathcal{X}$$

preserves and reflects limits, and in particular products.

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  • $\begingroup$ @Todd: Do you cite that M is closed under composition for (E,M)-structured categories when you want to explain someone that the composition of two injective maps is injective? ;) (Do not take this comment too seriously) $\endgroup$
    – Martin Brandenburg
    Commented Nov 27, 2011 at 21:07
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    $\begingroup$ Okay, I won't. :-) It's just another point of view, and may carry some resonance for some readers. If it doesn't, then fine. $\endgroup$
    – Todd Trimble
    Commented Nov 27, 2011 at 22:05
  • $\begingroup$ A bit more serious: Your proof only works if $X$ has coproducts. But the fact that $A/X \to X$ creates and preserves limits holds in general. If $X$ has coproducts, I would just use that $A \coprod -$ is left adjoint to $A/X \to X$. $\endgroup$
    – Martin Brandenburg
    Commented Nov 28, 2011 at 8:24
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    $\begingroup$ Yes, I know. But sometimes a little light can be shed with a mild extra hypothesis. Knowing that co-slices are categories of algebras, and that slices are categories of coalgebras, can come in handy on occasion, even if you have to assume a little extra hypothesis. That's really the reason for giving my answer -- to share a factoid that could come in handy for someone one day. I hope you don't mind! $\endgroup$
    – Todd Trimble
    Commented Nov 28, 2011 at 12:17

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