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So if $\mathcal{C}$ is locally small, we have the Yoneda embedding $Y:\mathcal{C} \rightarrow [\mathcal{C}^{op},Sets ]$. This preserves all limits in $\mathcal{C}$, and a comment here: An application of Yoneda Lemma says that showing this is just a matter of reformulating the definition of a limit. How is this?

I thought about how I would show that $Y$ preserves limits, and all I could think of was this: Suppose $L$ is the limit in $\mathcal{C}$ of some diagram $D$. Take some cone over $YD$ in $[\mathcal{C}^{op},Sets]$ with base $P$ and write $P$ as a colimit of representable presheaves. (I think) you can then use the Yoneda embedding to show that $L$ is a cocone over the elements of $P$, and the unique map $P \rightarrow YL$ from the universal property of $P$ will also work to show that $YL$ is indeed the limit.

Is this proof correct, and what is the easy proof that I've missed?

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If $\{L \to D_i\}$ is a limit diagram in $C$, then for every $X \in C$ we have that $\{\hom(X,L) \to \hom(X,D_i)\}_i$ is a limit diagram in $\mathsf{Set}$, which means precisely that $\{\hom(-,L) \to \hom(-,D_i)\}_i$ is a limit diagram in $[C^{op},\mathsf{Set}]$.

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  • $\begingroup$ Thanks. The fact that I didn't see that means I need to go understand the Yoneda lemma better, I suppose... $\endgroup$ – Zach L. Jun 19 '13 at 14:38
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    $\begingroup$ You only need the definition of $Y$, not anything else. $\endgroup$ – Martin Brandenburg Jun 19 '13 at 17:53
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    $\begingroup$ For a categorical novice, it might be worthwhile to explicitly state that the last step requires not just that $\{\hom(X,L) \to \hom(X,D_i)\}_i$ is a limit diagram, but that for each $X$, but also that the morphisms in these diagrams are natural in $X$. $\endgroup$ – Omar Antolín-Camarena Jun 19 '13 at 18:39
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    $\begingroup$ Yes this is the existence of $Y$. $\endgroup$ – Martin Brandenburg Jun 19 '13 at 19:42
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    $\begingroup$ There is no such functor $C \to Set^C$. Do you mean $C^{op} \to Set^C$? This is continuous by simply apply what has been proven to $C^{op}$ (since $Set^{(C^{op})^{op}}=Set^C$). $\endgroup$ – Martin Brandenburg Feb 10 '16 at 17:40

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