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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?

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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.

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  • $\begingroup$ How does this differ from what I said? Isn't what I said just this but with symbols? $\endgroup$ – Anonymous Jan 5 '20 at 7:50
  • $\begingroup$ For one I never mention the functor $\Delta$. $\endgroup$ – Connor Malin Jan 5 '20 at 9:30
  • $\begingroup$ 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. $\endgroup$ – Anonymous Jan 5 '20 at 11:36

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