Statement about Homotopy in Brown's “Topology & Groupoids”

I am trying to understand a statement in Brown's Topology and Groupoids, 7.2.5 (Corollary 1), page 270.

Let's first have some preliminary remarks

Let $X,Y$ be topological spaces. The track groupoid $\pi Y^X$ is defined as the groupoid whose objects are the maps $f:X\to Y$ and morphism in $\pi Y^X(f,g)$ are the homotopies $f\simeq g,$ where two homotopies $F,G$ are identified if they can be continuously deformed into each other while fixing the end maps. For $X=\{*\}$ we obtain the fundamental groupoid $\pi Y.$ A map $i:A\to X$ induces a morphism $i^*:\pi Y^X\to\pi Y^A$ by $i^*([F])=[F(i\times 1)].$

Let $i:A\to X,\ u:A\to Y.$ By $[(X,i),(Y,u)]$ we mean the set of homotopy classes of maps $f:X\to Y,fi=u,$ where a homotopy $F:f\simeq g$ is required to satisfy $F(ia,t)=ua.$ (We say $f$ and $g$ are homotopic under $i.$)

(You might want to skip this shaded part as it's probably not relevant for the particular question.):

Let $p : E → B$ be a morphism of groupoids. We say $p$ is a fibration if the following condition holds: for all objects $x$ of $E$ and elements $b$ in $B$ with initial point $px,$ there is an element $e$ of $E$ with initial point $x$ and such that $pe = b.$

If $p : E → B$ is a morphism of groupoids and $u$ is an object of $B,$ we write $p^{-1}[u]$ for the subgroupoid of $E$ with objects those $x$ in Ob$(E)$ such that $px = u,$ and with elements those $e$ in $E$ such that $pe = 1_u.$

Let $\pi_0G$ denote the set of components of $G.$ then we have:
Lemma 1: If $p:E\to B$ is a fibration of groupoids, then for each $b\in B(u,v)$ there is a bijection $$b_\#:\pi_0 p^{-1}[u]\to\pi_0 p^{-1}[v]$$ which respects identities and composition. Namely, if $x$ is an object with $px=u$, we assign to the component of $x$ the component of the object $y$ which is the codomain of the element $e$ such that $pe=b.$

Lemma 2: A cofibration $i:A\hookrightarrow X$ induces a fibration $i^*:\pi Y^X\to\pi Y^A.$

Lemma 3: Let $i:A\hookrightarrow X$ be a cofibration. Let $u:A\to Y$ be a map. Let $p=i^∗:πY^X\to πY^A.$ Then there is a canonical bijection $$π_0 p^{−1}[u]\cong[(X, i), (Y, u)].$$

In the proof of the Lemma 3 we use the following fact. Maybe it turns out to be useful in solving my question:

Lemma 4: Let $i:A→X$ be a cofibration. Let $H:X\times\Bbb I→Y$ be a homotopy $f≃g,$ and let $G=H(i\times1)$ be homotopic rel end maps to $G′:u≃v.$ Then $H$ is homotopic rel end maps to some $H′:f≃g$ such that $H'(i\times 1)=G'.$

Now we can combine the three lemmas to derive the following

Corollary: Let $i:A\hookrightarrow X$ be a cofibration and $α\in πY^A(u,v).$ Then there is a bijection $$α_\#:[(X, i), (Y, u)] → [(X, i), (Y, v)]$$

I have figured out that this bijection works as follows:
Choose a representative $F$ of $\alpha.$ For a homotopy class $[f]$ take a representative $f:X\to Y,\ fi=u.$ The homotopy $F:u\simeq v$ on $A$ can be extended to a homotopy $G:f\simeq g'.$ Then we have $[g']=\alpha_\#([f]).$

Now, Brown writes

Also, if $α_\#([f]) = [g],$ then any representative of $α$ extends to a homotopy $f\simeq g.$

I don't see why this should be obvious. I know that if $F:u\simeq v,\ F\in\alpha,$ then there is an extension of $f\cup F$ to a homotopy $G:f\simeq g'$ for some $g:X\to Y,\ g'i=v.$ And this $g'$ is homotopic under $i$ to $g.$ But I don't see how these homotopies can be combined to a homotopy $f\simeq g$ which extends $F.$

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Let $i:A\rightarrow X$ be a cofibration and $Y$ a topological space. With 7.2.5 (Lemma) on page 269 we find:

$\text{cls }H\left(i\times1\right)=\left\{ H'\left(i\times1\right)\mid H'\in\text{cls }H\right\}$ for every homotopy $H:X\times\mathbb{I}\rightarrow Y$.

Let $F:u\simeq v$ for $u,v:A\rightarrow Y$ and $fi=u$.

Let $\alpha:u\rightarrow v$ be an arrow in $\pi Y^{A}$ represented by $F$ (i.e. $\alpha=\text{cls }F$) and let $g$ be a representative of $\alpha_{\#}\left[f\right]$ here (i.e. $\alpha_{\#}\left[f\right]=\left[g\right]$).

Since $i^{*}:\pi Y^{X}\rightarrow\pi Y^{A}$ is a fibration there is an arrow $\text{cls }H:f\rightarrow g$ in $\pi Y^{X}$ with $i^{*}\left(\text{cls }H\right)=\text{cls }F=\alpha$. So $\alpha=\text{cls }H\left(i\times1\right)=\left\{ H'\left(i\times1\right)\mid H'\in\text{cls }H\right\}$ wich means that $F=H'\left(i\times1\right)$ for some $H':f\simeq g$

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I added an answer to my question, if you are interested. – Stefan Hamcke Oct 14 '13 at 21:18
I think your answer wasn't that trivial as I first thought. I shall think about it one more time. You are already using the lemma that I consider the key point of my answer now. – Stefan Hamcke Oct 14 '13 at 21:29
I think what you wrote has substance, but you could have added a few more details. For example, you could have written that the $g$ in your post is the $g$ from the assumption that $α_♯[f]=[g]$ (where $α=[F]$). And then you correctly conclude that any representative $H:f≃g$ of the lifting arrow $[H]$ has a restriction $H(i×1)$ which is homotopic rel $u,v$ to $F$, and then by the lemma $H$ is homotopic to an extension $H'$ of $F$ rel $f,g$. I didn't understand that back then, but now it's clear to me. If you could add that small detail to your answer, I'm willing to accept it ;-) – Stefan Hamcke Oct 15 '13 at 0:46
Well, I will dive into it again (it has been fading away in the meanwhile, I am getting older). Not at once (no time) but as soon as I have the opportunity. Thank you for this second chance. I will also have a good look at your answer. – drhab Oct 15 '13 at 7:22
I hope this is enough. If not then I am willing to adapt or add. – drhab Oct 16 '13 at 13:03

The lemma 3 is deeper than it looks : in there you have to prove that if there is an homotopy $G : f \to g$ such that $i^*(G) = F : u \to u$ is homotopic to $id_u$, then there is an homotopy $G' : f \to g$ that is above $i$ and $u$.

From an homotopy $H : F \to id_u$ and a pullback $G : f \to g$ of $F$, you can pullback $H$ to an homotopy $H' : G \to G'$ such that $i^*(H') = H$. Hence $i^*(G') = id_u$ and so $G' : f \to g$ is an homotopy above $i$ and $u$.

And this is actually used when showing that the homotopy class of $\alpha_\sharp(f)$ is well-defined by your construction :

Suppose $F : u \to v$ and $f : X \to Y$ such that $i^*(f) = u$.

By lemma 2, $i^* : \pi Y^X \to \pi Y^A$ is a fibration, and $F$ can be pulled back : there is some $g \in i^{*-1}(v)$ and some $G : f \to g$ such that $F = i^*(G)$. We pick $F_\sharp(f) = g$.

By lemma 1, if there is an homotopy between $f$ and $f'$ above $i,u$, then there is an homotopy between $g$ and $g'$ above $i,v$. In particular (picking $id_f : f \to f$), if there are several possible $g$ for the same $f$, they are homotopic : if $G : f \to g$ and $G' : f \to g'$ are two pullbacks of $F$, then $G' G^{-1} : g \to g'$ is an homotopy whose image by $i^*$ is $F F^{-1}$. This is not $id_v$ so this homotopy is not above $i$ and $v$. However, $F F^{-1}$ is homotopic to $id_v$. As it should be explained in lemma 3, you can pullback this starting from $G'G^{-1}$ to obtain a new homotopy $g \to g'$ whose image by $i^*$ is $id_v$, which means it is above $i$ and $v$.

Hence even though $g = F_\sharp(f)$ itself is not fully determined, its homotopy class is, and depends only on the homotopy class of $f$.

What's left is to show that if we pick another $F'$ homotopic to $F$ then $g'$ is homotopic to $g$ above $i$ and $v$, (and in fact, $G$ is homotopic to $G'$)

You can prove straight from lemma 3 that if $\alpha = [id_u]$ then $\alpha_\sharp([f]) = [f]$ forall $[f] \in [(X,u),(Y,u)]$ :

If $H : F \to id_u$ in $\pi Y^A(u,u)$ and if you have a pullback $G : f \to g$ of $F$, this is an actual arrow in $i^{*-1}(u)$ (since $i^*G = F \equiv id_u$). Hence $f$ and $g$ are in the same component in $\pi_0i^{*-1}(u)$. By lemma 3, they are in the same homotopy class above $i$ and $u$, and so $[f] = [g]$ (again, we pullback the homotopy between $F$ and $id_U$ starting from $G$ to get a new homotopy from $f$ to $g$ which is above $i$ and $u$)

Hence $id_{u\sharp} = id_{[(X,u),(Y,u)]}$. You can then argue that $F_\sharp G_\sharp = (FG)_\sharp$ and $(F_\sharp)^{-1} = (F^{-1})_\sharp$, then use the fact that if $F$ and $G$ are homotopic, then $G^{-1}F$ is homotopic to $id_U$, and conclude from there that $F_\sharp = G_\sharp$.

You can also do this by doing an actual pullback straight from the homotopy :

Suppose $F_1,F_2 : u \to v$, $H : F_1 \to F_2$ is an homotopy and $f : X \to Y$ satisfies $i^*(f) = u$.

Then we can pullback $F_1$ to an homotopy $G_1 : f \to g$ where $i^*(g) = v$ and $i^*(G_1) = F_1$. Then we can pullback $H$ to an homotopy $H' : G_1 \to G_2$ where $i^*(H') = H$ hence $i^*(G_2) = F_2$. So $G_2 : f' \to g'$ where $i^*(f') = u$ and $i^*(g') = v$.
The faces of $H'$ give you an homotopy $f \to f'$ above $i$ and $u$, an homotopy $g \to g'$ above $i$ and $v$. Combining those with $G_1$ or $G_2$ give homotopies between $f$ and $g$ (you can have one above $F_1$ and one above $F_2$), and they are homotopic (because $H'$ is the homotopy between them)

So we have a well-defined map between homotopy classes of homotopies $F : u \to v$ and homotopy classes of $f$ above $u$, to homotopy classes of $g$ above $v$, and in fact to homotopy classes of homotopies $G : f \to g$,

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This is (and will probably stay) beyond my scope. Brown calls it an 'immediate' consequence of the lemmas. How to explain that? – drhab Sep 24 '13 at 12:32
@drhab : Suppose $\alpha_\sharp([f]) = [g]$. Given a representative $F$ of $\alpha$ and $f$ of $[f]$ you can extend $F$ to an homotopy $G$ between $f$ and some $g' \in [g]$. Now, compose this with an homotopy above $i,v$ from $g'$ to $g$. Since the only thing that happens happens outside $A$, you can deform the resulting homotopy into one that still extends $F$. It's not exactly immediate though. I think he simply meant to say that you can find an extension of $F$ from $f$ to some $g'$ that is in the class of $g$. – mercio Sep 24 '13 at 13:26
Although it's a bit late, I wanted to thank you for your answer. I don't remember if I worked through it when I first read it. It must have taken a lot time to elaborate it. It does not answer the particular question I had, on the other hand, the solution is in what I've now added as "Lemma 4" to my post and which was not available to you when you wrote your answer. See my answer below if you're interested. – Stefan Hamcke Oct 14 '13 at 20:39

I have finally figured it out. The solution is to look carefully at the proof of Lemma 1 and its implications.

Assume that $b\in B(u,v)$. Let $x$ be an object in $p^{-1}[u]$, that is $px=u$. As $p$ is a fibration, there is an arrow $e$, starting at $x$ and ending at an object which we will denote $y$, such that $pe=b$. Then we can define $b_\#(\bar x)=\bar y$. Of course, we have to check this is well-defined. To this end, let $x'$ be another object in $\bar x$, and let $e'$ be an arrow in $E(x',y')$ with $pe'=b$. Then there is a morphism $d:x\to x'$ such that $pd=1_u$. Now, $e'de^{-1}$ is an arrow from $y$ to $y'$ and $p(e'de^{-1})=b1_yb^{-1}=1_v$. That means that $y$ and $y'$ are in the same component of $p^{-1}[v]$. So the action of $B$ on $\pi_0p^{-1}[u]$ is well-defined. $\square$

Additionally, if $b_\#(\bar x)=\bar y$, then there is an arrow $e$, starting at $x$ and ending at some $y'$ such that $pe=b$ and $\bar y'=\bar y$. But then there is also an arrow $d:y'\to y$ with $pd=1_v$. The composition $de$ maps to $b$. So we can say that $b_\#(\bar x)=\bar y$ if and only if some arrow from $x$ to $y$ lifts $b$.

Now, if $i:A↪X$ is a cofibration and $u:A→Y$ a map, then $p:=i^*:πY^X→πY^A$ is a fibration. Further, $πY^A$ operates on the family of sets $π_0 p^{-1}[u],\ u:A\to Y$ via $α_\#\bar x=\bar y$.
To answer the question, assume that $α_\#\bar f=\bar g$ and that $H:u≃v$ is an element in $α$. Then there is an arrow $β∈πY^X(f,g)$ which maps to $α$ via $p$. If $K:f≃g$ is a representative of $\beta$, then $pβ=α$ means that $K(i×1)$ is homotopic rel $u,v$ to $H.$ The lemma 4 in my question then asserts that $K$ is homotopic rel $f,g$ to a homotopy $K':f≃g$ which extends $H$. This shows that if $α_\#\bar f=\bar g$, any representative $H$ of $α$ can be extended to a homotopy $f≃g$.

In particular, we can take $α=1_u$ represented by the homotopy $H$ of lenght $r$ such that $H(a,s)=u(s)$. Then $α_\#(\bar f)=\bar g$ iff $f$ and $g$ are in the same component of $p^{-1}[u]$. By the above result the constant homotopy $H:u≃u$ can be extended to a homotopy $K:f≃g$, which means that $f$ and $g$ are homotopic under $A$ and thus they are in the same homotopy class of $[(X,i),(Y,u)]$. That's why we have a bijection $$π_0p^{−1}[u]≅[(X,i),(Y,u)]$$ and, consequently, $πY^A$ acts on the set of homotopy classes: $$α_\#:[(X,i),(Y,u)]\to[(X,i),(Y,v)]$$

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