# Why are limits defined by terminal morphisms?

Wikipedia page on universal properties gives an explanation of how limits are terminal morphisms:

Let $$F:J\to C$$ be a diagram in a category $$C$$. Moreover, let $$\Delta:C\to C^J$$ be the diagonal functor, which maps an object $$X$$ to the functor that maps all objects in $$J$$ to $$X$$, and all morphisms to $$id_X$$.

Then the limit of $$F$$ is a terminal morphism from $$\Delta$$ to $$F$$.

I am confused about this for the following reason: A terminal morphism from $$\Delta$$ to $$F$$ is a morphism in the category $$C^J$$, and thus a natural transformation. Let’s take $$J$$ to be the category with 2 objects and no non-identity morphisms, so that the limit is the product. Then a natural transformation here (morphism in $$C^J$$), will consist of a pair of two morphisms in $$C$$. Let $$(L,p_1,p_2)$$ be a candidate for the limit of $$F$$.

• It seems to me that for $$(L,p_1,p_2)$$ (considered as a natural transformation) to be a terminal morphism, we need that for any other $$(L’,p_1’,p_2’)$$, we need a unique natural transformation, i.e. a morphism in $$C^J$$, a pair of morphisms in $$C$$ $$f:L’\to L,g:L’\to L$$ such that $$p_1’=p_1\circ f$$ and $$p_2’=p_2\circ g$$, since morphisms in this context are natural transformations between objects in $$C^J$$.

• On the other hand, the actual definition of limit requires there to be a unique single morphism in $$C$$ (not a natural transformation, i.e. a single morphism in $$C^J$$) $$h:L’\to L$$, such that $$p_1’=p_1\circ h$$ and $$p_2’=p_2\circ h$$.

In particular, it seems to me that the first requirement implies that the product of the diagram in Set containing $$X,X$$ is $$X$$ itself, rather than $$X\times X$$.

Where is my thinking wrong?

• But a pair IS a single morphism in your example functor category. – Randall May 25 at 12:50
• @Randall, I was referring to a single morphism in $C$ – user56834 May 25 at 14:04

You seem confused about the definition of a terminal morphism. A terminal morphism from $$\Delta$$ to $$F$$ is a terminal object in the comma category $$(\Delta\downarrow F)$$. In particular, then, the unique morphism which is required to exist is a morphism in this comma category, not in $$C^J$$. An object of $$(\Delta\downarrow F)$$ is a pair $$(L,p)$$ where $$L$$ is an object of $$C$$ (the domain of $$\Delta$$) and $$p:\Delta(L)\to F$$ is a morphism in $$C^J$$. Given two such objects $$(L',p')$$ and $$(L,p)$$, a morphism between them is a morphism $$f:L'\to L$$ in $$C$$ (not in $$C^J$$) such that $$p\Delta(f)=p'$$. So, in order for $$(L,p)$$ to be terminal, the unique morphism that must exist is a morphism of $$C$$, not of $$C^J$$.
• Ah I see now. The reason I got confused is: the morphism that has to be unique is a morphism in $C$, but that morphism is still mapped to a natural transformation in $C^J$, and it is this natural transformation that is composed with $(p_1,p_2)$. It’s just that it’s a natural transformation consisting of a pair $(f,f)$ of identical morphisms in $C$. (Hence the name diagonal functor). – user56834 May 26 at 5:56