Let $X,X',Y,Y'$ be topological spaces (for the sake of readability I do not write the topologies) and suppose we are given $$ f: X\to Y\qquad\text{continuous},\\ f': X'\to Y'\qquad\text{continuous},\\ g:X\to X'\qquad\text{homotopy equivalence},\\ g':X'\to X\qquad\text{homotopy inverse of } g,\\ h:Y\to Y'\qquad\text{homotopy equivalence},\\ h':Y'\to Y\qquad\text{homotopy inverse of } h,\\ f'\circ g=h\circ f. $$ Is it true that $Y/f(X)\simeq Y'/f'(X')$?
Here is what I have done:
Let $\pi:Y\to Y/f(X)$ and $\pi':Y'\to Y'/f'(X')$ be the canonical (continuous) surjections. Using $$ Y\overset{h}{\longrightarrow}Y'\overset{\pi'}{\longrightarrow}Y'/f'(X'),\\ Y'\overset{h'}{\longrightarrow}Y\overset{\pi}{\longrightarrow}Y/f(X) $$ I have constructed, using also the extra condition $f'\circ g=h\circ f$, continuous maps $\varphi:Y/f(X)\to Y'/f'(X')$ and $\varphi':Y'/f'(X')\to Y/f(X)$ such that $$ \varphi\circ\pi=\pi'\circ h,\\ \varphi'\circ\pi'=\pi\circ h' $$ which then yields $$ \varphi'\circ\varphi\circ\pi=\pi\circ h'\circ h,\\ \varphi\circ\varphi'\circ\pi'=\pi'\circ h\circ h'. $$ But know I'm having trouble defining a homotopy for $\varphi'\circ\varphi$ resp. $\varphi\circ\varphi'$. Because if say $H:Y\times[0,1]\to Y$ is such that $H(y,0)=y, H(y,1)=h'(h(y))$ for all $y\in Y$, then it is very tempting to define $$ A:Y/f(X)\times[0,1]\to Y/f(X)\\ \pi(y)\mapsto \pi(H(y,t)), $$ but this might be ill defined because there is no way to assure that this doesn't depend on the representative of the class chosen. Can we surpass that problem? Also, I haven't used yet that $g$ is a homotopy equivalence which is kind of suspicious, but I don't see how to use it.
And last but not least, is it even true that $Y/f(X)\simeq Y'/f'(X')$? I naively assume it is, because I cannot think of a straightforward way to construct a counterexample.