1
$\begingroup$

Consider a simplicial category $\cal D$, and a cosimplicial object $c\colon \Delta\to \cal D$. Let $y\colon \Delta\to \bf sSet$ denote the Yoneda embedding.

How do you characterize the limit of $c$ weighted by $y$ (provided it exists)?

Partial answer: Whatever it is such an object,its associated representable presheaf must behave in such a way that $$ {\cal D}\big(d,\textstyle \text{lim}^yc\big)\cong {\bf Set}^{\Delta^\text{op}\times\Delta}\big( y,{\cal D}(d,c-) \big) $$ or in other words $$\begin{align*} {\cal D}\big(d, \text{lim}^yc\big) &\cong {\bf Set}^{\Delta^\text{op}\times\Delta}\big( y,{\cal D}(d,c-) \big)\\ & \int_{n\in\Delta}{\bf sSet}\big( \Delta_n, {\cal D}(d, c_n) \big)\\ &\cong \int_{n\in\Delta} {\cal D}(d, c_n)_n \end{align*}$$ which is the "cotrace" (?) of the profunctor $\Delta^\text{op}\times\Delta\to{\bf Sets}\colon (m,n)\mapsto {\cal D}(d, c_n)_m$. Is this something meaningful? How can it be generalized to the case of a functor $F\colon \cal C\to D$, where $\cal D$ is a $\widehat{\cal C}$-category?

$\endgroup$
4
  • $\begingroup$ I'm even more motivated now that I strongly suspect that the particular weighted limit I'm looking for is in some sense dual to the geometric realization of a simplicial set. Any idea is welcome! $\endgroup$
    – fosco
    Oct 9, 2013 at 5:18
  • $\begingroup$ It's called the "totalization" of your cosimplicial object and it is, indeed, dual to the geometric realization. $\endgroup$ Oct 9, 2013 at 20:30
  • 1
    $\begingroup$ Thanks! Can you give me a reference or answer in detail with some situations where such a thing appears? $\endgroup$
    – fosco
    Oct 9, 2013 at 21:35
  • $\begingroup$ There's a definition in 19.8.1 of Hirschorn's book on model categories, but it may not be very penetrable. In the case D=sSet, it's mentioned in section VII.5 of Goerss+Jardine's book Simplicial Homotopy Theory. $\endgroup$ Oct 10, 2013 at 22:28

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.