Consider the following commutative diagram. $A$, $B$, $C$, $D$ are CW-complexes. $a$ and $b$ are homotopy equivalences.
$\require{AMScd}$
\begin{CD}
A @>f>> B\\
@VaVV @VVbV\\
C @>>g> D
\end{CD}
Question: If $f$ and $g$ are fibrations, are their fibers homotopy equivalent?
Moreover, does $f^{-1}x\stackrel{a}{\longrightarrow}g^{-1}bx$ for any $x\in B$ give such a homotopy equivalence?
Coquestion: If $f$ and $g$ are cofibrations, are their cofibers homotopy equivalent?
Moreover, does $B/fA\longrightarrow D/gaA$ induced by $fA\stackrel{b}{\longrightarrow}gaA$ give such a homotopy equivalence?
They look like very elementary results. Could someone help me?
Edit I realized that they are just immediate consequences of Puppy sequences and the Whitehead theorem.
Best Answer
Thinking about it again, it might be slightly tricky. The spaces need not be CW-complexes. Now, consider the diagram $$\require{AMScd} \begin{CD} A @>{i}>> B @>>> C(i) @>>> B/A\\ @V{a}VV @V{b}VV @V{C(b,a)}VV @V{\overline{b}}VV\\ C @>{j}>> D @>>> C(j) @>>> D/C \end{CD}. $$ The maps $i,j$ are cofibrations and I am going to treat them as inclusions WLOG. The maps $a,b$ are homotopy equivalences. The somewhat tricky fact from the theory of cofibrations we have to call upon here is that $(b,a)\colon(B,A)\rightarrow(D,C)$ is then automatically a homotopy equivalence of pairs. Thus, we can find maps $c\colon C\rightarrow A$ and $d\colon D\rightarrow B$ such that $(d,c)\colon(D,C)\rightarrow(B,A)$ is a homotopy inverse. Pick a homotopy $H\colon B\times I\rightarrow B$ from $db\colon B\rightarrow B$ to $\mathrm{id}_B$ that restricts to a homotopy $A\times I\rightarrow A$ from $ca\colon A\rightarrow A$ to $\mathrm{id}_A$. Then, we obtain a homotopy $C(i)\times I\rightarrow C(i)$ from $C(d,c)C(b,a)$ to $\mathrm{id}_{C(i)}$ by mapping $(y,t)\mapsto H(y,t)$ for $y\in B$ and $([x,s],t)\mapsto[H(x,t),s]$ for $x\in A$. Analogously, we obtain a homotopy $C(b,a)C(d,c)\Rightarrow\mathrm{id}_{C(j)}$. Thus, $C(b,a)$ is a homotopy equivalence. Now, consider the right-most square in the diagram. The horizontal maps are both homotopy equivalences because $i$ and $j$ respectively are cofibrations and $C(b,a)$ is a homotopy equivalence by the argument just given, hence $\overline{b}$ is a homotopy equivalence.