# What is a codomain of diagonal functor?

I'm reading a "Graph Transformations. An Introduction to the Categorical Approach" by H.J.Schneider. In a example 6.3.3 Graph Category constructed as a comma category of a identity functor $id_{Set} : Set \rightarrow Set$ and a diagonal functor $\Delta_{Set} : Set \rightarrow Set \times Set$. The functors has a different codomains and couldn't be used to construct a comma category. Does it all mean that a $Set \times Set$ category is a subcategory of $Set$? And that comma category doesn't require equality of the functor codomains? Codomain of the one functor may be a subcategory of a codomain of the another functor?

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Diagonal functor from exercise 6.3.3 and diagonal functor from nlab are different things. In the example the codomain of diagonal functor is $\mathbf{Set}$, not $\mathbf{Set}\times\mathbf{Set}$. This functor sends every set $X$ to the set $X\times X\in\mathbf{Set}$, not to the pair $(X,X)\in\mathbf{Set}\times\mathbf{Set}$. – Oskar Jan 14 '14 at 9:02
Thanks! You dispelled my doubts :) But it's strange that a different things has a same name... – Denis Jan 14 '14 at 9:11
@Denis It's not strange: it's an unavoidable fact that there are only finitely many short words in any language... – Zhen Lin Jan 14 '14 at 9:15
@Denis I would not call it strange, I would rather say that it is unfortunate and unnecessary for an author to mis-use a well established functor name to represent something much less useful like his X functor. In his defence , at least the author clearly defines what he means and uses the letter X to hint at a cross product. – magma Jan 14 '14 at 14:17

$$\mathbf{Graph} = \mathbf{Set}^{\Gamma}$$
where $\Gamma$ is the category with exactly two objects and two distinct, parallel, non-identity arrows. In other words: A graph is a functor $G:\Gamma \rightarrow \mathbf{Set}$ and the natural transformations between these functors are the graph transformations. You can read this in Awodey (chap 7 I think).