Section 11. (0MTP) · ScalingStacks

11. Equivalences in Segal spaces[0MTP]

In this section we give a proof of (6.2). We use the Reedy model category structure in what follows.

We make use of an explicit filtration of $E=\discnerve(I[1])$ . Note that the category $I[1]$ has two objects, which we call $x$ and $y$ , and exactly four morphisms: $x\rightarrow x$ , $x\rightarrow y$ , $y\rightarrow x$ , $y\rightarrow y$ . Thus the morphisms are in one-to-one correspondence with the “words” $xx$ , $xy$ , $yx$ , $yy$ . In general the points of $E_{k}$ are in one-to-one correspondence with words of length $k+1$ in the letters $\{x,y\}$ . The “non-degenerate” points correspond to the words which alternate the letters $x$ and $y$ ; there are exactly two such non-degenerate points in $E_{k}$ for each $k$ .

We define a filtration

F(1)\approx E^{(1)}\subseteq E^{(2)}\subseteq E^{(3)}\subseteq\dots\subseteq E

of $E$ where $E^{(k)}$ is the smallest subobject containing the word $xyxyx\cdots$ of length $(k+1)$ . Note that $E=\bigcup_{k}E^{(k)}$ , and so $\Map_{s{\operatorname{\mathcal{S}}}}(E,W)\approx\lim_{n}\Map{s{\operatorname{\mathcal{S}}}}(E^{(n)},W)$ . We will prove (6.2) by actually proving the following stronger result.

[05VM]

Proposition 11.1. If $W$ is a Segal space and $n\geq 3$ , the map $\Map_{s{\operatorname{\mathcal{S}}}}(E^{(n)},W)\rightarrow W_{1}$ factors through the subspace $W_{\hoequiv}\subseteq W_{1}$ , and induces a weak equivalence $\Map_{s{\operatorname{\mathcal{S}}}}(E^{(n)},W)\rightarrow W_{\hoequiv}$ .

We prove (11.1) in (11.7).

[05VN]

Proof of (6.2) from (11.1). Since $E$ is the colimit of the $E^{(n)}$ along a sequence of cofibrations, it follows by (11.1) that $\Map_{s{\operatorname{\mathcal{S}}}}(E,W)$ is the inverse limit of the $\Map_{s{\operatorname{\mathcal{S}}}}(E^{(n)},W)$ along a tower of trivial fibrations. The proposition follows. ∎

11.2. Morphisms induced by compositions[0MTQ]

Let $W$ be a Segal space. Given $g\in\map(y,z)$ , consider the zig-zag

\map(x,y)\xrightarrow{\{g\}\times 1}\map(y,z)\times\map(x,y)\xleftarrow[\sim]{\varphi_{2}}\map(x,y,z)\xrightarrow{d_{1}}\map(x,z);

this induces a morphism $g_{*}\colon\map(x,y)\rightarrow\map(x,z)$ in the homotopy category of spaces. Likewise, given $f\in\map(x,y)$ , consider the zig-zag

\map(y,z)\xrightarrow{1\times\{f\}}\map(y,z)\times\map(x,y)\xleftarrow[\sim]{\varphi_{2}}\map(x,y,z)\xrightarrow{d_{1}}\map(x,z);

this induces a morphism $f^{*}\colon\map(y,z)\rightarrow\map(x,z)$ in the homotopy category of spaces. Note that if $f\in\map(x,y)$ and $g\in\map(y,z)$ , then $g_{*}([f])=f^{*}(g)=[g\circ f]$ (using the notation of §5). We have the following.

[05VP]

Proposition 11.3.

(1)

Given $f\in\map(x,y)$ and $g\in\map(y,z)$ , and $g\circ f$ the result of a composition, then $(g\circ f)_{*}\sim g_{*}\circ f_{*}$ and $(g\circ f)^{*}\sim f^{*}\circ g^{*}$ .
(2)

Given $x\in{\operatorname{ob}}W$ then $(\id_{x})_{*}\sim(\id_{x})^{*}\sim\id_{\map(x,x)}$ .

[05VQ]

Proof. To prove (1), let $k\in\map(x,y,z)$ be a composition of $f$ and $g$ which results in a composite $g\circ f$ . To show that $(g\circ f)_{*}\sim g_{*}\circ f_{*}$ , it suffices to show that both sides of the equation are equal (in the homotopy category of spaces) to the zig-zag

\map(w,x)\xrightarrow{\{k\}\times 1}\map(x,y,z)\times\map(w,x)\xleftarrow{\sim}\map(w,x,y,z)\rightarrow\map(w,z).

The proof that $(g\circ f)^{*}\sim f^{*}\circ g^{*}$ is similar.

The proof of (2) is straightforward. ∎

[05VR]

Proposition 11.4. Let $f,g\in\map(x,y)$ . Then $f\sim g$ if and only if the maps $f_{*},g_{*}\colon\map(w,x)\rightarrow\map(w,y)$ are homotopic for all $w\in{\operatorname{ob}}W$ , if and only if the maps $f^{*},g^{*}\colon\map(y,z)\rightarrow\map(x,z)$ are homotopic for all $z\in{\operatorname{ob}}W$ .

[05VS]

Proof. The only if direction is straightforward. To prove the if direction, suppose that $f_{*}$ and $g_{*}$ are homotopic for all $w\in{\operatorname{ob}}W$ . Then in particular they are homotopic for $w=x$ . The following commutative diagram demonstrates that $f_{*}(\id_{x})\sim f$ .

Similarly $g_{*}(\id_{x})\sim g$ , whence $f\sim g$ using (11.3), as desired. ∎

[05VT]

Corollary 11.5. If $f\in\map(x,y)$ is a homotopy equivalence (in the sense of (5.5)) then $f_{*}$ and $f^{*}$ are weak equivalences of spaces.

It is convenient to write $\map(x,y)_{f}$ to denote the component of $\map(x,y)$ containing $f$ . More generally, we write $\map(x_{0},\dots,x_{k})_{f_{1},\dots,f_{k}}$ for the component of $\map(x_{0},\dots,x_{k})$ corresponding to the component of $(f_{1},\dots,f_{k})$ in $\map(x_{0},x_{1})\times\dots\times\map(x_{k-1},x_{k})$ . The following lemma will be used in the proof of (11.1).

[05VU]

Lemma 11.6. Given a Segal space $W$ and $f\in\map(x,y)$ and $g\in\map(y,z)$ such that $f$ is a homotopy equivalence, the induced map

\map(x,y,z)_{f,g}\xrightarrow{(d_{1},d_{2})}\map(x,z)_{g\circ f}\times\map(x,y)_{f}

is a weak equivalence.

[05VV]

Proof. This follows from the diagram

Here the vertical column is a weak equivalence since $f$ is a homotopy equivalence (restricting to the fiber over $f\in\map(x,y)_{f}$ of the projections to $\map(x,y)_{f}$ gives exactly the zig-zag which defines $f^{*}\colon\map(y,z)_{g}\rightarrow\map(x,z)_{g\circ f}$ ). Since $(d_{0},d_{2})$ is a weak equivalence, the lemma follows. ∎

11.7. Proof of (11.1)[0MTR]

For $k\geq 2$ there are push-out diagrams

(11.8)

where $\sigma_{k}$ is the map corresponding to the word $xyx\cdots$ of length $(k+1)$ , and where $H(k)$ denotes the largest subobject of $F(k)$ not containing $d_{0}\iota$ .

We next note that $H(k)$ can itself be decomposed. Thus let $C(k)\subseteq F(k)$ denote the largest subobject of $F(k)$ not containing $d_{0}d_{0}\iota$ . If we let $d^{1}\colon F(k-1)\rightarrow F(k)$ denote the inclusion of the “face” $d_{1}\iota$ , then we have that $d^{1}F(k-1)\cap C(k)=d^{1}H(k-1)$ , and thus an isomorphism

H(k)\approx C(k)\cup_{d^{1}H(k-1)}d^{1}F(k-1).

(11.9)

Let $X$ be a simplicial space and $W$ a Segal space. Then each map $\gamma\colon F(1)\rightarrow X$ induces a map

\gamma^{*}\colon\Map_{s{\operatorname{\mathcal{S}}}}(X,W)\rightarrow\Map_{s{\operatorname{\mathcal{S}}}}(F(1),W)\approx W_{1}

of spaces. We introduce the following notation. Let $\Map_{s{\operatorname{\mathcal{S}}}}(X,W)_{\hoequiv}$ denote the subspace of $\Map_{s{\operatorname{\mathcal{S}}}}(X,W)$ consisting of all simplices $x$ such that $\gamma^{*}(x)\in W_{\hoequiv}\subset W_{1}$ for all $\gamma\colon F(1)\rightarrow X$ . Then $\Map_{s{\operatorname{\mathcal{S}}}}(X,W)_{\hoequiv}$ is isomorphic to a union of some of the path components of $\Map_{s{\operatorname{\mathcal{S}}}}(X,W)$ . In particular, $\Map_{s{\operatorname{\mathcal{S}}}}(F(1),W)_{\hoequiv}\approx W_{\hoequiv}$ by definition, and so $\Map_{s{\operatorname{\mathcal{S}}}}(F(k),W)_{\hoequiv}\approx W_{\hoequiv}\times_{W_{0}}\dots\times_{W_{0}}W_{\hoequiv}$ .

[05VW]

Lemma 11.10. Let $W$ be a Segal space. Then for $k\geq 2$ the induced map

\Map_{s{\operatorname{\mathcal{S}}}}(F(k),W)_{\hoequiv}\rightarrow\Map_{s{\operatorname{\mathcal{S}}}}(H(k),W)_{\hoequiv}

is a weak equivalence.

[05VX]

Proof. The proof is by induction on $k$ . The case $k=2$ is immediate from (11.6).

Now suppose the lemma is proved for the map $\Map_{s{\operatorname{\mathcal{S}}}}(F(k-1),W)_{\hoequiv}\rightarrow\Map_{s{\operatorname{\mathcal{S}}}}(H(k-1),W)_{\hoequiv}$ . From (11.9) we get a commutative square

This square would be a pullback square if we left off the “ $\hoequiv$ ” decorations. Even with these decorations the square is a pullback (and hence a homotopy pullback), as can be seen by recalling that $H(k)_{1}=C(k)_{1}\cup d^{1}F(k-1)_{1}$ .

Thus by induction we see that the map

a\colon\Map_{s{\operatorname{\mathcal{S}}}}(H(k),W)_{\hoequiv}\rightarrow\Map_{s{\operatorname{\mathcal{S}}}}(C(k),W)_{\hoequiv}

is a weak equivalence. The proof now follows from (11.11) and the fact that the map

W_{k}\approx W_{k-1}\times_{W_{0}}W_{1}\xrightarrow{a\times_{W_{0}}1}\Map_{s{\operatorname{\mathcal{S}}}}(C(k),W)\approx\Map_{s{\operatorname{\mathcal{S}}}}(d^{1}H(k-1),W)\times_{W_{0}}W_{1}

is a weak equivalence after restricting to the “ $\hoequiv$ ” components. ∎

[05VY]

Lemma 11.11. There is a natural weak equivalence

\Map_{s{\operatorname{\mathcal{S}}}}(C(k),W)\approx\Map(d^{1}H(k-1),W)\times_{W_{0}}W_{1}.

[05VZ]

Proof. Let $d^{0}H(k-1)\subset C(k)$ denote the image of $H(k-1)$ in $C(k)$ induced by the map $d^{0}\colon F(k-1)\rightarrow F(k)$ . There is a square

of subobjects of $C(k)$ ; we need to show that the inclusion map $d^{0}H(k-1)\cup\alpha^{0}F(1)\rightarrow C(k)$ of the union of these subobjects is a weak equivalence in the Segal space model category structure.

Now $C(k)$ can be written as a colimit of the poset of subcomplexes each of which

(1)

are isomorphic to $F(\ell)$ for some $\ell<k$ , and
(2)

include $0,1\in F(k)_{0}$ .

Straightforward calculation shows that the intersection of $d^{0}H(k-1)\cup\alpha^{0}F(1)$ with each of the objects $F(\ell)$ in the above diagram is a cover of $F(\ell)$ . ∎

[05W0]

Proof of (11.1). It is clear that for $k\geq 3$ every map

\Map_{s{\operatorname{\mathcal{S}}}}(E^{(k)},W)\rightarrow\Map_{s{\operatorname{\mathcal{S}}}}(F(1),W)\approx W_{1}

induced by an inclusion $F(1)\rightarrow E^{(k)}$ must factor through $W_{\hoequiv}\subseteq W_{1}$ , since each point of the mapping space maps to a homotopy equivalence in the sense of (5.5). Let $r_{k}$ denote the map $\Map_{s{\operatorname{\mathcal{S}}}}(E^{(k)},W)\rightarrow W_{1}$ associated to the inclusion $F(1)\rightarrow E^{(k)}$ classifying the point $xy\in E^{(k)}_{1}$ . We have that $\Map_{s{\operatorname{\mathcal{S}}}}(E^{(k)},W)=\Map_{s{\operatorname{\mathcal{S}}}}(E^{(k)},W)_{\hoequiv}$ for $k\geq 3$ , and even when $k=2$ we have that

\Map_{s{\operatorname{\mathcal{S}}}}(E^{(2)},W)_{\hoequiv}\approx\Map_{s{\operatorname{\mathcal{S}}}}(E^{(2)},W)\times_{W_{1}}W_{\hoequiv}.

Then we must show that for each $k\geq 2$ the fiber of $r_{k}$ over any point in the subspace $W_{\hoequiv}\subset W_{1}$ is contractible. The result now follows from (11.10) applied to the pushout diagrams (11.8). ∎

Charles Rezk

Original source: arXiv:math/9811037v3

Original source · math/9811037v3