A homotopy equivalence between spaces $B\Gamma$ and $K\Gamma$ for a graph of groups on a graph $\Gamma$

In Hatcher is "Algebraic Topology" (p. 92), the room $B\Gamma$ (for a chart of teams on a chart $\Gamma$) is specified to be a collection of rooms $BG_v$ for each and every vertex $v$, which are attached by particular mapping cyndrical tubes representing the side morphisms. If we change $BG_v$ with any kind of $K(G_v,1)$ room in the building and construction, call the resulting room $K\Gamma$. After that there is a comment:

"We leave it to the viewers to examine that the resulting room $K\Gamma$ is homotopy equal to the room $B\Gamma$ created above."

My inquiry is: just how does one confirm this?

What I've obtained: I can not appear to expand the homotopy equivalences $BG_v\to K(G_v,1)$ to a homotopy equivalence of the whole network of mapping cyndrical tubes. I intend it suffices to do it for the instance of just one side as long as both homotopies that control the homotopy equivalence are constant on both sides of both mapping cyndrical tubes.

Offered a team morphism $G_v\to G_w$, I can see that the adhering to layout commutes upto homotopy:

$BG_v\longrightarrow BG_w$

$\downarrow\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \downarrow$

$K(G_v,1)\to K(G_w,1)$

given that both make-ups of maps generate the very same map $\pi_1(BG_v)\to\pi_1(K(G_w,1))$. (See Proposition 1B.9 on web page 90). Yet I do not recognize where to go from there.

Update : What I'm seeking is Hatcher is planned remedy, which possibly entails clearly contructing a homotopy equivalence.

The inquiry comes down to this. Intend we have a commutative layout


$A\longrightarrow B$

$\downarrow f\;\;\;\;\;\;\;\;\downarrow g$

$C\longrightarrow D$


where $f$ and also $g$ are homotopy equivalences, $f^{-1}$ and also $g^{-1}$ are their homotopy inverses. Intend additionally that the layout is homotopy commutative therefore is the layout with the homotopy inverses. Allow $M_1$ be the mapping cyndrical tube for $f_1$ and also $M_2$ be the mapping cyndrical tube for $f_2$.

Intend we have the adhering to homotopies:

  • $H_1:A\times I\to D$, a homotopy $f_2 f \simeq g f_1$
  • $H_2:C\times I\to B$, a homotopy $f_1 f^{-1} \simeq g^{-1}f_2$
  • $H_f:A\times I\to A$, a homotopy $f^{-1}f\simeq 1_A$
  • $H_g:B\times I\to B$, a homotopy $g^{-1}g\simeq 1_B$ ¢ ¢ Define $F:M_1\to M_2$ to be:
  • $(f(a),2t)$ for $(a,t)\in A\times [0,\frac{1}{2}]$
  • $H_1(a,2t-1)$ for $(a,t)\in A\times [\frac{1}{2}.1]$
  • $g(b)$ for $b\in B$
  • ¢ ¢ Define $G:M_2\to M_1$ to be:

  • $(f^{-1}(c),2t)$ for $(c,t)\in C\times [0,\frac{1}{2}]$
  • $H_2(c,2t-1)$ for $(c,t)\in C\times [\frac{1}{2},1]$
  • $g^{-1}(d)$ for $d\in D$
  • ¢ ¢ We intend to show that $GF\simeq 1$ making use of a homotopy that expands both $H_f$ and also $H_g$ ¢ ¢ My effort:

    Computing $GF$ offers:

  • $(f^{-1}f(a),4t)$ for $(a,t)\in A\times[0,\frac{1}{4}]$
  • $H_2(f(a),4t-1)$ for $(a,t)\in A\times[\frac{1}{4},\frac{1}{2}]$
  • $g^{-1}(H_1(a,2t-1))$ for $(a,t)\in A\times[\frac{1}{2},1]$
  • $g^{-1}g(b)$ for $b\in B$
  • ¢ ¢ We can partly specify $H:M_1\times I\to M_1$ as adheres to:

  • $(H_f(a,s),4(1-s)t+st)$ for $(a,t,s)\in A\times [0,\frac{1}{4}]\times I$
  • For $(a,t,s)\in A\times [\frac{1}{4},\frac{1}{2}]\times I$ as adheres to:
  • $H_2(f(a), \frac{1-4t}{2t-\frac{1}{2}}s + 4t-1)$ for $s\leq 2t-\frac{1}{2}$
  • $(H_f(a, \frac{s-(2t-\frac{1}{2})}{1-(2t-\frac{1}{2})}), t + \frac{(t-1)s-(t-1)}{1-(2t-\frac{1}{2})})$ for $s\geq 2t-\frac{1}{2}$
  • ¢ ¢ So what I require is a function $H:A\times[\frac{1}{2},1]\times I\to M_1$ that pleases the adhering to border problems:

  • Agrees with the currently specified $H$ on $A\times\{\frac{1}{2}\}\times I$
  • Is equivalent to $g^{-1}(H_1(a,2t-1))$ on $A\times[\frac{1}{2},1]\times\{0\}$
  • Is equivalent to $(a,t)$ on $A\times[\frac{1}{2},1]\times\{1\}$
  • Is equivalent to $H_g(f_1(a),s)$ on $A\times \{1\}\times I$
  • 5
    2022-06-07 14:35:29
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    Answers: 1

    You get on the appropriate track, I think.

    You intend to show that a homotopy commutative layout $$ \begin{array}{ccc} X_1&\to&Y_1\\ \downarrow\psi & &\downarrow\phi\\ X_2&\to&Y_2. \end{array} $$ includes a (purely) commutative layout $$ \begin{array}{ccccc} X_1&\to&\operatorname{Cyl}(f_1)&\gets&Y_1\\ \downarrow& &\downarrow\pi&&\downarrow\\ X_2&\to&\operatorname{Cyl}(f_2)&\gets&Y_2. \end{array} $$ Let $F\colon X_1\times I\to Y_2$ be the homotopy from the first layout. Simply specify $\pi$ to be

    • $\psi\times(t\mapsto 2t)$ on $X_1\times[0;1/2]\subset\operatorname{Cyl}(f_1)$ ;
    • $F$ on $X_1\times [1/2;1)\subset\operatorname{Cyl}(f_1)$ ;
    • $\phi$ on $Y_1\subset\operatorname{Cyl}(f_1)$

    Applying this lemma per side generates a map $f\colon B\Gamma\to K\Gamma$.

    Allow is show that $f$ without a doubt is a homotopy equivalence. Confirming homotopy equivalence straight is generally tough, so allow is use algebraic approaches rather.

    Represent by $V_1\subset B\Gamma$ (resp. $V_2\subset K\Gamma$) the (disjoint) union of rooms in vertices. Observe that both $B\Gamma/V_1$ and also $K\Gamma/V_2$ are of the very same homotopy type $$ E:=\bigcup_{\text{edge}\in\Gamma}\Sigma K(G_{\operatorname{source}(\text{edge})},1)\cong\Gamma\vee\bigvee_{\text{edge}\in\Gamma}\Sigma K(G_{\operatorname{source}(\text{edge})},1). $$

    Now make believe for a minute that basic teams of all rooms are abelian. After that it suffices to show that $f$ generates isomorphism in homology. Yet it adheres to quickly from using the 5 - lemma to homology long specific series of sets $(B\Gamma,V_1)$ and also $(K\Gamma,V_2)$: $$ \begin{array}{ccccccccc} H_{i+1}(E_1)&\to&H_i(V_1)&\to&H_i(B\Gamma)&\to&H_i(E_1)&\to&H_{i-1}(V_1)\\ \wr| & &\wr| & &\downarrow f_*& &\wr| & & \wr|\\ H_{i+1}(E_2)&\to&H_i(V_2)&\to&H_i(K\Gamma)&\to&H_i(E_2)&\to&H_{i-1}(V_2). \end{array} $$ Now our rooms might have non - unimportant basic teams, yet revealing that $f$ generates isomorphism in homology is still sufficient if one makes use of homology with coefficients in neighborhood systems (oh, one requires to show first that basic teams are isomorphic bdsh that can be done by Seifert - van Kampen). So in fact we are done.

    (Morale of the tale: using 5 - lemma to homology long specific series permits to confirm points like "if $a_1\cong a_2$ and also $b_1\cong b_2$, after that $a_1+b_1\cong a_2+b_2$" for weak homotopy kinds, offered there is one map generating both weak equivalences.)

    2022-06-08 03:54:02