# Using Yoneda's Lemma to show that limits are unique

I recall my professor mentioning that Yoneda's Lemma can be used to show that limits of a functor are unique up to isomorphism. Here is my attempt:

Let $$F:J\to\mathcal{C}$$ be a functor and let $$X$$ and $$Y$$ be limits for the functor. Then for every object $$Z\in C$$, we have a bijection $$\text{Hom}_\mathcal{C}(Z,X)\cong \text{Cone}(Z,F)=\text{Hom}_{\text{Psh}(\mathcal{C})}(\Delta(Z),F)$$ where $$\text{Psh}(\mathcal{C})=\text{Fun}(\mathcal{C}^{op},\textbf{Set})$$ and $$\Delta(Z):J\to\mathcal{C}$$ is the constant functor whose image is $$Z$$. These bijections together constitute a natural isomorphism from $$\text{Hom}_\mathcal{C}(\cdot\ ,X)$$ to $$\text{Cone}(\cdot\ ,F)$$ as functors from $$\mathcal{C}^{op}$$ to $$\textbf{Set}$$. By the same reasoning, $$\text{Hom}_\mathcal{C}(\cdot\ ,Y)\cong\text{Cone}(\cdot\ ,F)$$ so that $$\text{Hom}_\mathcal{C}(\cdot\ ,X)\cong\text{Hom}_\mathcal{C}(\cdot\ ,Y)$$. However, because the Yoneda embedding is fully faithful (a consequence of the Yoneda Lemma), this implies $$X\cong Y$$ in $$\mathcal{C}$$.

Working through the isomorphisms, one can see that the isomorphism from $$X$$ to $$Y$$ obtained here is actually the morphism induced by the universality of $$Y$$, so the limit is actually unique up to a canonical isomorphism with respect to the limiting cone. Is all my reasoning correct?

## 1 Answer

Your reasoning is correct, but you can stop after the first natural isomorphism. You are stating that since $$X$$ is a limit it represents the functor sending an object to the set of cones with that object as its summit. Symmetrically, $$Y$$ also represents this functor. By the Yoneda lemma, these objects must be isomorphic through an isomorphism respecting the representations.

• How does this differ from what I said? Isn't what I said just this but with symbols? – Anonymous Jan 5 '20 at 7:50
• For one I never mention the functor $\Delta$. – Connor Malin Jan 5 '20 at 9:30
• But you mention cones, which are effectively defined using the functor $\Delta$ (or in some equivalent manner). I don't think your comment is substantially different from my attempt. – Anonymous Jan 5 '20 at 11:36