# Category of Elements and Terminal Object

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.

• The category of elements is just the slice category $\mathscr{C}/C$. Apr 22 '21 at 17:47
• I can only think of one “natural” object... is it terminal? Apr 22 '21 at 18:16
• 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? Apr 22 '21 at 21:52
• @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. Apr 22 '21 at 22:19
• I wouldn’t write “$f(y)=x$“ (which suggests a set function taking values), but rather $x=y\circ f$ (indicating composition). Apr 22 '21 at 23:11

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 $$\int\mathscr{C}(-,C)$$.

• 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. Apr 22 '21 at 23:48