Suppose that $0\to G_1\to G_0\to A\to 0$ is a free resolution for $A$ ($A$ is an Abelian group) and $0\to G_1'\to G_0'\to A\to 0$ is another free resolution for $A$ so I have built the morphisms $f_0: G_0\to G_0'$ and $f_1:G_1\to G_1'$ and what I need is to prove that the morphisms $f_0, f_1$ are unique except homotopies of chain complexes.
$$\require{AMScd} \begin{CD} 0 @> >> G_1 @> \alpha >> G_0 @> \beta >> A @>>>0 \\ @. @V f_1 VV @V f_0 VV @V Id VV @. \\ 0 @> >> G_1' @> \alpha' >> G_0' @> \beta' >> A @> >>0 \end{CD} $$
To do this I am taking other $g_0:G_0\to G_0'$, $g_1:G_1\to G_1'$ morphisms and my intention is to prove that $f_*$ and $g_*$ are homotopic, to do this I must build a homotopy $T_*$ such that $T_0:A\to G_0'$, $T_1:G_0\to G_1'$ and
$\beta'T_0=Id\\ \alpha'T_1+T_0\beta=f_0-g_0\\ T_1\alpha=f_1-g_1$
I think that $T_0$ can be easily constructed because $\beta':G_0'\to A$ is surjective and so there is an inverse right $T_0:A\to G_0'$ such that $\beta'T_0=Id$.
To construct $T_1:G_0\to G_1'$ we take a base $\{a_n\}$ for $G_0$ and define $T_1(a_n)=b_n$ so that $\alpha'(b_n)=(f_0-g_0)(a_n)$, such $b_n$ exists because $\beta'((f_0-g_0)(a_n))=\beta'f_0(a_n)-\beta'g_0(a_n)=\beta(a_n)-\beta(a_n)=0$ and so $(f_0-g_0)(a_n)\in ker(\beta')=Im(\alpha')$
This is where I'm stuck, I don't know if what I've done is fine and I don't know how to prove that $\alpha'T_1+T_0\beta=f_0-g_0$ and that $T_1\alpha=f_1-g_1$, could someone please help me?