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I am dealing with the following problem:

Consider a locally small category $\mathscr{C}$ and consider $C\in\mathscr{C}$. What are the objects and morphisms in the category of elements of $\mathscr{C}(-,C)$? Also, find a terminal object of $\int \mathscr{C}(-,C)$.

I think the objects of $\int\mathscr{C}(-,C)$ are the pairs $(D,x)$ where $D\in\mathscr{C}$ and $x\in \mathscr{C}(D,C)$, and the morphisms of $\int\mathscr{C}(-,C)$ are the maps $f:(D,x)\to(E,y)$ where $f:D\to E$ in $\mathscr{C}$ such that $\mathscr{C}(-,C)f(y)=x$. I can't really find the terminal object though.
Any insight would be helpful! Thanks in advance.

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    $\begingroup$ The category of elements is just the slice category $\mathscr{C}/C$. $\endgroup$ Apr 22, 2021 at 17:47
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    $\begingroup$ I can only think of one “natural” object... is it terminal? $\endgroup$ Apr 22, 2021 at 18:16
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    $\begingroup$ Just try and see! Suppose you have an object $(D, x)$. Is there at least one morphism $(D, x) \to (C, \textrm{id}_C)$? Is it forced to be something specific? $\endgroup$
    – Zhen Lin
    Apr 22, 2021 at 21:52
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    $\begingroup$ @SummerAtlas Yes, I second Zhen Lin. Write down what it means for $f:D\to C$ to be a morphism $(D,x)\to(C,id_C)$ and it should be clear that there is a unique $f$ that does the job. $\endgroup$ Apr 22, 2021 at 22:19
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    $\begingroup$ I wouldn’t write “$f(y)=x$“ (which suggests a set function taking values), but rather $x=y\circ f$ (indicating composition). $\endgroup$ Apr 22, 2021 at 23:11

1 Answer 1

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So here's the solution I figured out from the hints in the comments.

We work on the contravariant category for the problem.

The objects in the category of elements of $\mathscr{C}(-,C)$ is the collection of pairs $(D,x)$ where $D\in\mathscr{C}$ and $x\in \mathscr{C}(D,C)$. The morphisms in the category of elements of $\mathscr{C}(-,C)$ is the collection of maps $f:(D,x)\to(E,y)$ where $f:D\to E$ in $\mathscr{C}$ such that $\mathscr{C}(-,C)f(y)=x$, which is equivalent to $y\circ f = x$ by definition of Hom functor.

We claim that $(C,\text{id}_C)$ is a terminal object of $\int \mathscr{C}(-,C)$. Consider arbitrary $(D,x)\in \int\mathscr{C}(-,C)$. For starter, we show there exists a morphism $f:(D,x)\to (C,\text{id}_C)$ in $\mathscr{C}(-,C)$. Observe that $D\in\mathscr{C}$ and $x\in\mathscr{C}(D,C)$. Then by definition $x:D\to C$ is a morphism in $\mathscr{C}$. Furthermore, $\mathscr{C}(-,C)x(\text{id}_C) = \text{id}_C\circ x = x$ by definition of Hom functor. Then by definition we obtain a morphism $f:(D,x)\to (C,\text{id}_C)$ in $\int\mathscr{C}(-,C)$.

We now show the uniqueness of morphism from $(D,x)$ to $(C,\text{id}_C)$. Indeed, for any two morphisms $f,g:(D,x)\to (C,\text{id}_C)$ in the category of elements, we know $f:D\to C$ and $g:D\to C$ are morphisms in $\mathscr{C}$, and $\text{id}_C\circ f=x=\text{id}_C\circ g$. Therefore $f=g$. Hence, the morphism $f:(D,x)\to (C,\text{id}_C)$ is unique.

Concluding from the properties above, $(C,\text{id}_C)$ is a terminal object of $\int\mathscr{C}(-,C)$.

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    $\begingroup$ Looks good. But you don't need to make a contradiction. Just show that every morphism from $(D,x)$ to $(C,\mathrm{id}_C)$ is equal to $x : D \to C$. A direct proof is just one line. Remember for other proofs as well that almost always a proof of contradiction should not be the first choice, since it overcomplicates things. $\endgroup$ Apr 22, 2021 at 23:48

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