$Y / f(X)$ and $Y' / f'(X')$ are homotopy equivalent? My Question is:
Let $f : X \rightarrow Y$ and $f' : X' \rightarrow Y'$ two continuous maps.
Let $g: X \rightarrow X'$ and $h: Y \rightarrow Y'$ two homotopy equivalences
such that $f'g = hf$.
Is it true that $Y/f(X)$ and $Y'/f'(X')$ are homotopy equivalent?
I tought a lot but I don't manage to show something.
My idea was first to show that $f(X) \simeq f'(X')$. And then that the quotients are homotopy equivalent. 
Then I searched a counterexample but without sucess because of the condition $f'g = hf$.
Thank you for your help. (Just a small tip that I can try the rest by myself is welcome too.)
 A: Firstly, you should replace the naive quotients $Y/f(X)$ and $Y'/f'(X')$ by the homotopy cofibers $C_{f}$, $C_{f'}$, obtained by turning $f$, $f'$ into cofibrations. Otherwise the statement is clearly not true, For example, take $f:S^{n-1}\hookrightarrow D^n$ and $f':S^{n-1}\rightarrow \ast$.
With this replacement made, you can use the standard cofibres
$$C_f=\frac{X\times I\bigsqcup Y}{(x,1)\sim f(x)},\qquad C_{f'}=\frac{X'\times I\bigsqcup Y'}{(x',1)\sim f'(x')},$$
to explicitly write down an induced map $k:C_f\rightarrow C_{f'}$ and construct a homotopy inverse from the supplied data. You still need the general result, however, since you will want to show that the homotopy types of $C_f$, $C_{f'}$ are independent of any particular choices made in turning $f$, $f'$ into a cofibration.
Now the statement is true, but it is not immediate. If $X,Y,X'Y'$ are CW complexes, with $Y,Y'$ simply connected, then $C_f$, $C_{f'}$ are also CW, and you may use the Seifert-van Kampen theorem to show that $C_{f}$ and $C_{f'}$ are also simply connected. Studying the long exact sequences in homology, you apply the 5-lemma to conclude that $k$ induces isomorphisms $k_*:H_*C_f\xrightarrow{\cong}H_*C_{f'}$. Since these spaces are simply connected you get that $k$ is a weak equivalence. Since these spaces are CW you conclude that $k:C_f\xrightarrow{\simeq}C_{f'}$ is in fact a homotopy equivalence.
The general case for $Y,Y'$ simply connected follows from the special case by CW approximation. Of course, in this case you only conclude that the cofibres are weakly equivalent.
The statement still holds true in the case that $Y,Y'$ are not simply connected. The idea of the proof is the same, but there are more technical details. In particular you need to replace ordinary homology $H_*$ with homology $\mathcal{H}_*(-;\underline A)$ with local coefficients in an arbitrary $\pi_1Y$-module $\underline A$ (the induced maps make any $\pi_1Y'$-, $\pi_1C_f$-, or $\pi_1C_{f'}$-module into a $\pi_1Y$-module).
The Seifert-van Kampen theorem shows that $k_*:\pi_1C_f\xrightarrow{\cong}\pi_1C_{f'}$ is an isomorphism, and with this established you again study the long exact homology sequences, this time with local coefficients. You conclude again from the 5-lemma that $\mathcal{H}_*(C_f;\underline A)\xrightarrow{\cong}\mathcal{H}_*(C_{f'};\underline A)$ is an isomorphism for all $\pi_1C$-modules $\underline A$ and use the more detailed versions of the previous theorems to conclude that $k:C_f\xrightarrow{\simeq}C_{f'}$ is a homotopy equivalence.
