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The fact that $\mathscr A(A,-):\mathscr A\to \textbf{Set}$ preserves $D$-indexed limits translates to $$\lim\mathscr A(A,D(-))\simeq \mathscr A(A,\lim D)$$

I'm trying to prove that the dualized version should be $$\lim \mathscr A(D(-),A)\simeq \mathscr A(\text{colim} D,A)$$ (it's not immediately obvious to me why we should have colimit on the RHS and why the order flipped compared to the first displayed equation). So in words, the dual statement says that $\mathscr A(-,A):\mathscr A^{op}\to \textbf {Set}$ preserves $D$-indexed limits. But I'm having trouble unraveling this. For, suppose $D:I\to \mathscr A$ is a diagram of shape $I$ on $\mathscr A$. "Preserves $D$-indexed limits" should mean that the image of the limit of $D$ under the given functor is isomorphic to the limit of the composite diagram. But in this case, the composite diagram is not defined (the domain of $\mathscr A(-,A) $ is not equal to the range of $D$). Of course one can consider, instead of $D$, a diagram $D':I\to\mathscr A^{op}$, but I thought officially $D$-indexed limit preservation means that limits of diagrams $D:I\to \mathscr A$ are preserved, so nothing entitles us to consider that $D'$ diagram. So my question is how to deal with this matter?

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    $\begingroup$ Limit preservation means preservation of all limits --- for all $D$ and all $I$. If you want to require this only for a specific $D$, you'd say "preservation of $D$-indexed limits". $\endgroup$ May 12, 2021 at 23:55
  • $\begingroup$ ... except that $D$ here does not stand for an indexing category, but rather for a diagram. So the terminology doesn't fit anyway. $\endgroup$ May 15, 2021 at 2:41

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The statements say exactly the same, and it is not necessary to give a second proof. You know that if $D : I \to \mathcal{A}$ is a small diagram whose limit (in $\mathcal{A}$) exists, then $$\lim \mathcal{A}(A,D(-)) \cong \mathcal{A}(A,\lim(D)).$$ Hence, for any small diagram $D : I \to \mathcal{A}$, we can apply this to the dual diagram $D^{\mathrm{op}} : I^{\mathrm{op}} \to \mathcal{A}^{\mathrm{op}}$: If the limit of $D^{\mathrm{op}}$ exists (in $\mathcal{A}^{\mathrm{op}}$), then $$\lim \mathcal{A}^{\mathrm{op}}(A,D^{\mathrm{op}}(-)) \cong \mathcal{A}^{\mathrm{op}}(A,\lim(D^{\mathrm{op}}))).$$ Since $\mathcal{A}^{\mathrm{op}}(A,B) := \mathcal{A}(B,A)$, $D^{\mathrm{op}}(i) := D(i)$, and since we have $$\lim(D^{\mathrm{op}}) = \mathrm{colim}(D),$$ this can be written as $$\lim \mathcal{A}(D(-),A) \cong \mathcal{A}(\mathrm{colim}(D),A).$$

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The formula $\lim_i \mathscr A(Di,a) \cong\mathscr A(\text{colim}_i\, Di,a)$ is correct. It is in fact just a reformulation of the universal property of colimits. First note that the functor $\mathcal A(-,a)$ is contravariant. If you have a colimit cone $\mu: D \Rightarrow l$ in $\mathcal A$ then this colimit cone is in fact a limit cone in $\mathcal A^{op}$. Hence the formula $\lim_i \mathscr A(Di,a) \cong\mathscr A(\text{colim}_i\, Di,a)$ says that $\mathcal A(-,a)$ preserves limits. The switch from colimits to limits comes because we like to denote arrows in $\mathcal A^{op}$ by the corresponding arrows in $\mathcal A$. We often do not fully commit to the opposite category when we work in $\mathcal A$, because it makes things more complicated than they have to be. We just say, oh, that variable is contravariant, that one is covariant. I think this is where your confusion comes from. If you want to be precise, you need to consider $D$ as the functor $D: I^{op} \to \mathcal A^{op}$.

Next I will give you some intuition why the formula should be true. Here is a heuristic which works in many concrete categories: To give an element in a limit $x\in \lim_i Di$ is the same as giving elements $x_i\in Di$ from each object $Di$ such that a bunch of equations hold, one for each arrow in the category $I$. This is certainly true in $\mathbf{Set}$. Hence an element $f\in \lim_i\mathcal A(Di,a)$ is the same as a collection of elements $f_i \in \mathcal A(Di,a)$ such that a bunch of equations hold. Those are precisesly the equations which need to hold in order for $(f_i)$ to be a cocone under $D$. So an element of $\lim_i\mathcal A(Di,a)$ is, in a natural way, the same as a cocone under $D$. The next step is clear. The universal property of colimits states that a cocone under $D$ with apex $a$ is the same as an arrow $\text{colim} D \to a$. Hence an element of $\lim_i\mathcal A(Di,a)$ is the same as an element of $\mathcal A(\text{colim} D,a)$. This is the bijection we were looking for.

The explanation is longer than I intended. Here is the main point: The isomorphism $\lim_i \mathscr A(Di,a) \cong\mathscr A(\text{colim}_i\, Di,a)$ makes the heuristic precise that a map out of $\text{colim} D$ is the same as abunch of maps out of the components $Di$ which satisfy certain equations. Bonus: The following fact should now the unsurprising. If a cocone $\mu: D\Rightarrow l$ gets mapped to limit cones in $\textbf{Set}$ by all contravariant functors $\mathcal A(-,a)$, then it is a colimit. The premisse just means that $\mu$ satisfies the defining property of a colimit for all $a$.

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  • $\begingroup$ "We often do not fully commit to the opposite category, because it is to complicated." Well I do, and it is actually very simple. $\endgroup$ May 15, 2021 at 2:38
  • $\begingroup$ @Martin Brandenburg: I meant to say: often I am inside some category, say of R modules. I do something it it and it just happens that I have a colimit cone. Then I like to apply both contravariant and covariant functors to it to send it to other categories, and often I do not bother to first invert all the arrows. Hence the colimit in the formula. Is this a better explanation? $\endgroup$
    – Nico
    May 15, 2021 at 7:53
  • $\begingroup$ If we would view all contravariant functors truly as covariant functors from the opposite category, there would be no need to write first colimit and then limit. This is of course an artifact of that we started with a diagram in A instead of the its opposite. $\endgroup$
    – Nico
    May 15, 2021 at 8:06

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