$\text{Ext}^1(-,B)$, independence of projective resolution. If we want to compute the group $\text{Ext}^1(A,B)$ we take a projective resolution of $A$
$$\cdots\to P_2 \to P_1 \to P_0 \to A \to 0,$$ apply the contravariant functor $\text{Hom}(\cdot,B)$ to it (without $A$) and take 1st cohomology of the resulting chain complex. If we take another projective resolution then $\text{Ext}^1(A,B)$, even all $\text{Ext}^i$, remains isomorphic.
Why?  One argument resorts to saying that the two resolution are chain homotopic, but
I don't see why they are and how the isomorphism follows.  
As this is crucial part for the definition of Ext, I would like someone to shed some light on it.

P.J.Hilton, U.Stammbach, A Course in Homological Algebra: 
Corollary 3.5. If $\phi\cong\psi: C\to D$ and if $F$ is an additive functor, then 
$H(F\phi) = H(F\psi):H(FC)\to H(FD)$. 
