Remark 14.1. This completion is a generalization of the classifying space construction. In fact, suppose is a Segal space such that is a groupoid; equivalently, that . Then the arguments below show that is weakly equivalent to a constant simplicial space, which in each degree is the realization . For instance, if is a “-space” (i.e., ) and thus a model for a loop space with underlying space equivalent to , then is equivalent to the constant object which is , the classifying space of the “loop space” , in each degree.
14. A completion functor[0MTZ]
In this section we prove (7.7). We do this by constructing functorially for each Segal space a map called the completion map, such that
- (1)
the completion is a complete Segal space,
- (2)
the completion map is a weak equivalence in the complete Segal space model category, and
- (3)
the completion map is a Dwyer-Kan equivalence.
Statement (2) implies that a map between Segal spaces is a weak equivalence in the complete Segal space model category structure if and only if is. Likewise, statement (3) together with (7.5) imply that is a Dwyer-Kan equivalence if and only if is. Thus (7.7) will follow from statement (1) together with (7.6), which shows that the Dwyer-Kan equivalences between complete Segal spaces are precisely the Reedy weak equivalences between such, which are precisely the weak equivalences between fibrant objects in the complete Segal space model category structure.
We should note that it is easy to demonstrate statements (1) and (2) alone. In fact, (7.2) implies that there exists for each simplicial space a fibrant replacement map , in which is a weak equivalence in the complete Segal space model category structure, and is a complete Segal space. However, we need a different construction to prove all three statements.
Suppose is a Segal space. Let . For each we can define a simplicial space by . Let
Then the spaces taken together form a simplicial space , and there is a natural map . Since , we can write , where denotes the prolongation of the functor to simplicial objects in .
Let denote the functorial Reedy fibrant replacement of . The composite map is called the completion map of , and the functor which sends to is called the completion functor.
Lemma 14.2. If is a category, then is isomorphic to . In particular, and are weakly equivalent to the terminal object in .
Proof. The first statement is straightforward from the definitions. Since is equivalent to the terminal object in , and takes equivalences to weak equivalences by (3.7), the second statement follows. ∎
Lemma 14.3. If is a categorical equivalence between Segal spaces, then is a Reedy weak equivalence.
Proof. It is clear from the definition that , and is contractible by (14.2). Thus categorically homotopic maps are taken to homotopic maps by the completion operator, and hence completion takes categorical equivalences to homotopy equivalences. ∎
Proof of statements (1) and (3). For each simplicial map there is a diagram
By (13.8) the maps are Dwyer-Kan equivalences, so for each set of objects the morphism
between the fibers of the vertical maps in the above diagram is a weak equivalence. Thus, the square is a homotopy pullback, with fibers which are weakly equivalent to the products of mapping spaces.
Thus, the induced map of realizations has its homotopy fibers weakly equivalent to a -fold product of mapping spaces, and thus we have shown that is a Segal space, and that are weak equivalences for all .
By construction the map is surjective; it follows that is surjective on isomorphism classes of objects. Therefore we have shown that is a Dwyer-Kan equivalence, proving statement (3).
It remains to show that is a complete Segal space. Consider the square
induced by a map . Since is a categorical equivalence by (13.5) and thus a Dwyer-Kan equivalence by (13.8), we may conclude that the induced map is a weak equivalence for each pair . Thus the above square is a homotopy pullback, and so the induced map has its homotopy fibers weakly equivalent to the spaces . That is,
Since by (6.2), the above really says that there is an equivalence . Now (14.3) shows that since is categorically equivalent to , we have that ; in other words, is a complete Segal space. This proves statement (1), and completes the proof. ∎
Original source: arXiv:math/9811037v3
Original source · math/9811037v3