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In class we saw that given a diagram $D:\mathbb I\to \mathbb D$, one can define a contravariant functor $L:\mathbb D\to \mathbf {Set}$, which sends an object $X$ to the cones of $D$ with vertex $X$. Then, we proved that an object $Y$ is (the vertex of) the limit of $D$ iff $L$ is represented by $Y$.

However we also defined the functor $L$ more esplicitely: given an object $X$ in $\mathbb D$, $LX$ is the vertex of the limit of the diagram $Hom (X,D(-)):\mathbb I\to \mathbf {Set}$. My question is: this last way of viewing any $LX$ is necessary to prove the proposition above? It doesn't seem to me, but I'm quite new to category theory and I could easily be missing something. Thanks in advance

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    $\begingroup$ What is $H^X D$ ? I guess it is $Hom(X, D(-))$, but you should state your notations. $\endgroup$ – J. Darné May 4 at 8:18
  • $\begingroup$ Yes it's as you say $\endgroup$ – Dorian May 4 at 8:22
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    $\begingroup$ Ok, then you should edit your question to say so $\endgroup$ – J. Darné May 4 at 8:23
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You can prove the proposition without referring to the second point of view. But the second point of view, coupled with the proposition, is really the useful one. Namely, you will see it countless time, written as: $$Hom(X, lim(D)) = lim\ Hom(X, D)$$ (or, dually, $Hom(colim(D), X) = lim\ Hom(D, X)$). The term on the right here is $L(X)$ (using the second point of view), and "=" is in fact a natural isomorphism saying that $lim(D)$ represents $L(X)$. This completely caracterizes $lim(D)$, so you can take it as a definition of $lim(D)$, although it is less intuitive than speaking about cones of maps. But it is very useful for calculations. And I invite you to think about what this projective limit of sets is ; then you will see how these two points of view are equivalent.

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  • $\begingroup$ Thanks; so in order to prove the natural bijection that you wrote, starting from what I saw in class, I could just define $lim\ Hom(X,D)$ on arrows and prove that it is naturally isomorphic to the functor $L$ in my question (which should be easy immediate since the define the same thing). Is this right? $\endgroup$ – Dorian May 4 at 12:56
  • $\begingroup$ Yes, you should be able to see that they are the same thing (which was exactly the point of your second description, if I understand it correctly), and then a representative is unique (up to an iso). $\endgroup$ – J. Darné 2 days ago

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