Subsection 11.2. (0MTQ) · ScalingStacks

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. ∎

Charles Rezk

Original source: arXiv:math/9811037v3

Original source · math/9811037v3