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For a given weight $W : \mathcal{S}^{op} \to \mathcal{V}$ and diagram $D : \mathcal{S} \to \mathcal{A}$, the weighted colimit is an object $W \cdot D$ together with an isomorphism $$\mathcal{A}(W\cdot D, a) = \[\mathcal{S}^{op},\mathcal{V}\](W, \mathcal{A}(D-, a))$$ natural in $a$. If the weighted colimit exists, then it in turn induces a contravariant presheaf $$\mathcal{A}(D-, W\cdot D) : \mathcal{S}^{op} \to \mathcal{V}$$ with a natural transformation $$W \Rightarrow \mathcal{A}(D-, W\cdot D)$$ Generally, I'm interested in understanding in which situations (typically defined by a choice of diagram) the following questions have positive/negative answers:

  1. Is this natural transformation forced to be an iso?
  2. Do there exist weights $W_1, W_2 : \mathcal{S}^{op} \to V$ such that (the weighted colimits $W_1 \cdot D$ and $W_2\cdot D$ exist and) $$\mathcal{A}(D-, W_1\cdot D) \cong \mathcal{A}(D-, W_2\cdot D)$$ but nonetheless $W_1 \not\cong W_2$?

(Is there a diagram giving a positive answer to 1? This would imply a negative answer to 2, but is there also a (single) diagram giving a negative answer to both 1 and 2?)

More specifically, consider the case where $\mathcal{A}$ is the Eilenberg-Moore category of algebras for a monad, $\mathcal{S}$ the Kleisli category, and $D$ the inclusion functor. Since $D$ is dense, every algebra $a$ may be described as the $\mathcal{A}(D-,a)$ weighted colimit of $D$. Now, is it possible to find a different weight $W$ over $\mathcal{S}$, such that still we have $a = W\cdot D$?

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  • $\begingroup$ Let $\mathbb{V} = \mathbf{Set}$. 1) it is hardly an iso; take $W(X) = 1$; then $\mathbb{A}(D(-), W \cdot D) = \mathbb{A}(D(-), \mathit{colim}(D))$; you shall easily find an example, where $\mathbb{A}(D(-), \mathit{colim}(D)) \not\approx 1$. For 2) take $\mathbb{A} = 1$; for any diagram $D$ and any weight $W$, we have $W \cdot D = 1$. $\endgroup$
    – Michal R. Przybylek
    Commented Jul 18, 2012 at 18:14
  • $\begingroup$ Of course, for your more specific question, you may take the identity monad on the terminal category. $\endgroup$
    – Michal R. Przybylek
    Commented Jul 18, 2012 at 19:34
  • $\begingroup$ @Michal, thank you for your comment, but on the general questions I was hoping for more conceptual answers (e.g., when is the natural transformation forced to be an iso? clearly it doesn't have to be in general), if anyone wanted to offer them. I will revise the wording to try to make that more clear. On the specific question in the subject line, yes your example works (at least in the Set-enriched case), but it is (or at least it seems to me) a degenerate situation. Typically, I want to understand what happens when $\mathcal{A}(D-,a)$ itself is not trivial. $\endgroup$
    – Noam Zeilberger
    Commented Jul 18, 2012 at 20:29
  • $\begingroup$ @Noam, to give a specific example you always have to assume very specific things. However, the point, I was trying to make, is that there are no reasons for the natural transformation from your first question to be an isomorphism --- it is almost never an isomorphism (it may be an isomorphism in very degenerated cases). And the other point was that the situations where colimits over different diagrams or over different weights are isomorphic are really common. I suggest you to work out simple examples. $\endgroup$
    – Michal R. Przybylek
    Commented Jul 19, 2012 at 10:08
  • $\begingroup$ Also, I am not sure if I fully understand your specific question. Perhaps, what you would like to know is when the Elienberg-Moore category of a monad is the free $D$-cocompletion of its Kleisli category under some classes of weights. Am I right? $\endgroup$
    – Michal R. Przybylek
    Commented Jul 19, 2012 at 10:08

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