The Stacks project

Lemma 13.40.5. Let $\mathcal{D}$ be a triangulated category. Given a map (13.40.1.1) 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

\[ \xymatrix{ Y_0 \ar[r] \ar[d] & Y_1[1] \ar[r] \ar[ld] & Y_2[2] \ar[r] \ar[lld] & \ldots \ar[r] & Y_ n[n] \ar[lllld] \\ Y'_ n[n] } \]

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 $Y_0 \to Y_1[1] \to X_1[1]$ we see that $\mathop{\mathrm{Hom}}\nolimits (X_1[1], Y'_ n[n]) = 0$ guarantees that the second vertical arrow is unique. Since we have the distinguished triangle $Y_1[1] \to Y_2[2] \to X_2[2]$ we see that $\mathop{\mathrm{Hom}}\nolimits (X_2[2], Y'_ n[n]) = 0$ 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 13.40.4 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FXE. Beware of the difference between the letter 'O' and the digit '0'.