Product in undercategory 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!
 A: 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 $\mathrm{pr}_i f = f_i$.
A: 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. 
