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My recent answer to Is it possible to define higher cardinal arithmetics (about defining infinite tetrations) requires something I don't know. Here is the simplest case.

Take a set $S$ and consider $$ \mathbf{Set}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\dots $$ What is the colimit of this diagram?

I realize that formally there is an absolutely explicit construction and description of this colimit. What I want to know is whether the colimit is equivalent to some known manageable category. I don't even see whether it is equivalent to $\mathbf{Set}$.

The same for the (inverse) limit of the backwards diagram $$ \dots\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set}^{\mathrm{op}}\xrightarrow{\hom_{\mathbf{Set}}(\_,S)}\mathbf{Set} $$

I cannot even figure out whether the limit possesses an initial or a terminal object.

As a (probably) more lightweight version, - replace $\mathbf{Set}$ with the (large) groupoid of sets-and-bijections (but again take the functor $\hom_{\mathbf{Set}}$, rather than $\hom$ in this groupoid (since otherwise it would give something trivial)).

Update

Following suggestion of Philippe Gaucher I checked the simplest cases, with $S$ empty or singleton. In these cases both the limit and colimit are, respectively, $0\to1$ and the terminal category.

But I also realized that both the limit and colimit come out equivalent to their own opposites, so one never gets anything equivalent to $\mathbf{Set}$.

It is thus even more unclear what does one get in the limit for general $S$. Also I still do not see whether initial or terminal object exists there. Already the case of two element $S$ I don't see.

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    $\begingroup$ If $C(S)$ is your colimit and $L(S)$ your limit, did you try to calculate $C(\varnothing)$, $L(\varnothing)$, $C(S \times T)$, $L(S \times T)$, $C(\{0\})$, $L(\{0\})$ ? $\endgroup$ Aug 19, 2016 at 11:05
  • $\begingroup$ @Philippe Do you mean the functors $C:\mathbf{Set}\to\varinjlim{\cal D}$ and $L:\varprojlim{\cal D}'\to\mathbf{Set}$? If yes, then the problem is that the values of $C$ and arguments of $L$ live in unknown categories. So if e. g. $L(\varnothing)$ means value of $L$ on the initial object, I don't know whether $\varprojlim{\cal D}'$ has an initial object. On the other hand, I know that $C(\varnothing)$ exists but I don't know what it is - it is an object in $\varinjlim{\cal D}$, and I have no knowledge about that category. $\endgroup$ Aug 19, 2016 at 12:09
  • $\begingroup$ Oops sorry now I understand. Yes, good suggestion. $L(\varnothing)$ seems to have just two objects. Also $C(\varnothing)$, I think. For singletons, they seem to be single-morphism categories. $\endgroup$ Aug 19, 2016 at 12:15
  • $\begingroup$ For products, I don't know, have to think. $\endgroup$ Aug 19, 2016 at 12:21
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    $\begingroup$ Is it possible we get something like a Chu construction, $Chu(Set, S)$? (This is, by construction, equivalent to its own opposite.) $\endgroup$
    – Todd Trimble
    Aug 21, 2016 at 15:30

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