# Can we “complete” a category without products to have products?

Let $\mathcal{C}$ be a category without products, can we "complete" it into a category with products?

A presheaf is a contravariant functor from some topology to a category, in order to describe the condition of being a sheaf, we need $\mathcal{C}$ to have products. I am wondering if we can sheafify a general presheaf into a sheaf, but I am not sure how to define the concept of a sheaf in general case?

Consider the category of presheaves $[C^{op},Set]$ along with the yoneda embedding $Y : C \to [C^{op},Set]$. $[C^{op},Set]$ is closed under limits and colimits because $Set$ is, and they are computed pointwise. In particular, products exists.
• This certainly embeds $\mathcal C$ into a category with products, but "completion" should imply a universal such embedding, which this is not. – Kevin Carlson Oct 18 '15 at 15:55
• @KevinCarlson You are right that the category of presheaves is not the completion of $C$. However, it does have products and do exactly what the OP wants : define sheaves on $C$. More precisely, if $F$ is a presheaf on a site $C$, and $x\in C$, then by Yoneda $F(x)=Hom(h_x,F)$. The sheaf condition reads $F(x)=\operatorname{eq}(Hom(\prod h_{u_i},F)\rightrightarrows Hom(\prod h_{u_i}\times_{h_x} h_{u_j},F))$ for every covering $\{u_i\}$ of $x$. – Roland Oct 18 '15 at 16:02
• For the universal thing you actually want the free completion $[C, \text{Set}]^{op}$ rather than the free cocompletion. – Qiaochu Yuan Oct 18 '15 at 17:10
• @mqx: I'm not sure what Yuan had in mind, but for the natural embedding $C \to [C,Set]^{op}$ one has the continuous right Kan-extension $Ran_F : [C,Set]^{op} \to D$ along any functor $F : C \to D$ ($D$ complete), in particular it preserves all products. Which is something one might expect out of a universal completion. I agree that this is a more natural thing to do in this context. – Bubbles Oct 18 '15 at 19:06