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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?

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  • $\begingroup$ But a pair IS a single morphism in your example functor category. $\endgroup$
    – Randall
    May 25, 2019 at 12:50
  • $\begingroup$ @Randall, I was referring to a single morphism in $C$ $\endgroup$
    – user56834
    May 25, 2019 at 14:04

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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 $C^J$.

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  • $\begingroup$ 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). $\endgroup$
    – user56834
    May 26, 2019 at 5:56

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