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In chapter 6 on homotopy (co-)limits of Jeffrey Strom's Modern Classical Homotopy Theory he gives the following definition on p. 156 (I'm explaining the terminology at the bottom):

[Let $\mathscr{C}$ be a category, $I$ be a simple category with its filtration $I_0 \subseteq I_1 \subseteq \dots$ and $F:I \to \mathscr{C}$ be a functor.] We define $\widehat{F}_n : I_{n+1} \to \mathscr{C}$ by setting $\widehat{F}_n|_{I_n} = F_n$ and $\widehat{F}_{n+1}(i) = \operatorname{colim} F_{<i}$.

I don't get what Strom actually means here - the definition itself doesn't make sense to me. Is this maybe a typo and he means $\widehat{F}_{n}(i) = \operatorname{colim} F_{<i}$ for $i \in I_{n+1} \setminus I_n$? (There doesn't seem to be an errata for his book on the internet.)

The chapter itself focuses on the construction of cofibrant replacements for simple diagrams. Strom writes (p. 157) that $\widehat{F}_n$ is used to interpolate between $F_n$ and $F_{n+1}$ in an extension/lifting problem (hence the title of this question).


Let me explain the terminology here:

  • A simple category $I$ is a poset regarded as a category such that $$ d(i) = \sup \{n \mid \text{there is } x_0 \to \dots \to x_n = i \text{ of non-identity maps in } I \} < \infty $$ for every $i \in I$.
  • We take the full subcategory $I_n = \{i \in I \mid d(i) \leq n \}$ and write $F_n : I_n \to \mathscr{C}$ as the composition $I_n \to I \to \mathscr{C}$.
  • We take the full subcategory $I_{< i} = \{j \in I \mid j < i \}$ and write $F_{<i} : I_{<i} \to \mathscr{C}$ as the composition $I_{<i} \to I \to \mathscr{C}$.
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  • $\begingroup$ I suspect this is a special case of the general construction for Reedy categories. $\hat{F}_n$ is probably what is usually called the latching object. $\endgroup$
    – Zhen Lin
    Nov 25 at 14:39
  • $\begingroup$ @ZhenLin Thank you! I'm not familiar with Reedy categories (yet), but from glancing at nLab it indeed looks similar. However, I still can't really guess what $\widehat{F}_n$ should be. In particular, I can't find any similarities to $\widehat{F}_n|_{I_n} = F_n$. Do you have a guess? $\endgroup$
    – Qi Zhu
    Nov 26 at 8:47

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