Lemma 13.41.5. Let $\mathcal{D}$ be a triangulated category. Given a map (13.41.1.1) assume we are given Postnikov systems for both complexes. If

1. $\mathop{\mathrm{Hom}}\nolimits (X_ i[i], Y'_ n[n]) = 0$ for $i = 1, \ldots , n$, or

2. $\mathop{\mathrm{Hom}}\nolimits (Y_ n[n], X'_{n - i}[n - i]) = 0$ for $i = 1, \ldots , n$, or

3. $\mathop{\mathrm{Hom}}\nolimits (X_{j - i}[-i + 1], X'_ j) = 0$ and $\mathop{\mathrm{Hom}}\nolimits (X_ j, X'_{j - i}[-i]) = 0$ for $j \geq i > 0$,

then there exists at most one morphism between these Postnikov systems.

Proof. Proof of (1). Look at the following diagram

$\xymatrix{ Y_0 \ar[r] \ar[d] & Y_1 \ar[r] \ar[ld] & Y_2 \ar[r] \ar[lld] & \ldots \ar[r] & Y_ n[n] \ar[lllld] \\ Y'_ n[n] }$

The arrows are the composition of the morphism $Y_ n[n] \to Y'_ n[n]$ and the morphism $Y_ i[i] \to Y_ n[n]$. The arrow $Y_0 \to Y'_ n[n]$ is determined as it is the composition $Y_0 = X_0 \to X'_0 = Y'_0 \to Y'_ n[n]$. Since we have the distinguished triangle $Y_0 \to Y_1 \to X_1$ we see that $\mathop{\mathrm{Hom}}\nolimits (X_1, Y'_ n[n]) = 0$ guarantees that the second vertical arrow is unique. Since we have the distinguished triangle $Y_1 \to Y_2 \to X_2$ we see that $\mathop{\mathrm{Hom}}\nolimits (X_2, Y'_ n[n]) = 0$ guarantees that the third vertical arrow is unique. And so on.

Proof of (2). The composition $Y_ n[n] \to Y'_ n[n] \to X_ n[n]$ is the same as the composition $Y_ n[n] \to X_ n[n] \to X'_ n[n]$ and hence is unique. Then using the distinguished triangle $Y'_{n - 1}[n - 1] \to Y'_ n[n] \to X'_ n[n]$ we see that it suffices to show $\mathop{\mathrm{Hom}}\nolimits (Y_ n[n], Y'_{n - 1}[n - 1]) = 0$. Using the distinguished triangles

$Y'_{n - i - 1}[n - i - 1] \to Y'_{n - i}[n - i] \to X'_{n - i}[n - i]$

we get this vanishing from our assumption. Small details omitted.

Proof of (3). Looking at the proof of Lemma 13.41.4 and arguing by induction on $n$ it suffices to show that the dotted arrow in the morphism of triangles

$\xymatrix{ Y_{n - 1}[-1] \ar[d] \ar[r] & Y_ n \ar[r] \ar@{..>}[d] & X_ n \ar[r] \ar[d] & Y_{n - 1} \ar[d] \\ Y'_{n - 1}[-1] \ar[r] & Y'_ n \ar[r] & X'_ n \ar[r] & Y'_{n - 1} }$

is unique. By Lemma 13.4.8 part (5) it suffices to show that $\mathop{\mathrm{Hom}}\nolimits (Y_{n - 1}, X'_ n) = 0$ and $\mathop{\mathrm{Hom}}\nolimits (X_ n, Y'_{n - 1}[-1]) = 0$. To prove the first vanishing we use the distinguished triangles $Y_{n - i - 1}[-i] \to Y_{n - i}[-(i - 1)] \to X_{n - i}[-(i - 1)]$ for $i > 0$ and induction on $i$ to see that the assumed vanishing of $\mathop{\mathrm{Hom}}\nolimits (X_{n - i}[-i + 1], X'_ n)$ is enough. For the second we similarly use the distinguished triangles $Y'_{n - i - 1}[-i - 1] \to Y'_{n - i}[-i] \to X'_{n - i}[-i]$ to see that the assumed vanishing of $\mathop{\mathrm{Hom}}\nolimits (X_ n, X'_{n - i}[-i])$ is enough as well. $\square$

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