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In EGA I, Grothendieck writes (I'm paraphrasing):

Let $\mathbf{K}$ be a category, $(A_{\alpha})_{\alpha \in I}, (A_{\alpha \beta})_{(\alpha,\beta) \in I \times I}$ two families of objects of $\mathbf{K}$ for which $A_{\beta \alpha} = A_{\alpha \beta}$, $(\rho_{\alpha \beta})_{(\alpha,\beta) \in I \times I}$ a family of morphisms $\rho_{\alpha \beta}: A_{\alpha} \rightarrow A_{\alpha \beta}$. We will say that a pair consisting of an object $A$ of $\mathbf{K}$ and a family of morphisms $\rho_{\alpha}: A \rightarrow A_{\alpha}$ is a solution of the universal problem defined by the data of families $(A_{\alpha}), (A_{\alpha \beta})$, and $(\rho_{\alpha \beta})$ if the following holds:

For any object $B$, and any collection of morphisms $f_{\alpha}:B \rightarrow A_{\alpha}$ such that $\rho_{\alpha \beta} \circ f_{\alpha} = \rho_{\beta \alpha} \circ f_{\beta}$, there exists a unique morphism $f: B \rightarrow A$ such that $\rho_{\alpha} \circ f = f_{\alpha}$.

I understand the definition, but I don't understand how this is related to previous universal problems I've encountered. For example, the categorical product $X \times Y$ of two objects, together with projections maps $X \times Y \rightarrow X,Y$ satisfies a universal property, but I don't understand how it is related to the formulation here.

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    $\begingroup$ In general, it is impossible to rewrite the universal property of the product as such an universal problem. However, if the category has a terminal object $\ast$ then you get the universal property of the product by taking the universal problem with $A_{\alpha\beta} = \ast$. $\endgroup$ – Ben Jan 20 '16 at 17:40
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    $\begingroup$ After thinking about it for a while: this concept of universal problems is a common generalisation of a) fibre products and b) products indexed over sets. A solution can be written as a certain fibre product of products. Therefore, if the category has all pairwise fibre products and all products, there always exists a solution. The other way around, in a category where all universal problems have a solution, all products and fibre products exist. Since there is no clear question, I'm not sure whether this is the kind of response you (@D_S) want. Is it? $\endgroup$ – Ben Jan 21 '16 at 17:21

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