Homotopy equivalence of fibers and cofibers

Question

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?

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Edit I realized that they are just immediate consequences of Puppy sequences and the Whitehead theorem.

Answer 1

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.