Lemma 13.41.6. Let $\mathcal{D}$ be a triangulated category. Let $X_ n \to X_{n - 1} \to \ldots \to X_0$ be a complex in $\mathcal{D}$. If
then there exists a Postnikov system. If we have
then any two Postnikov systems are isomorphic.
Proof. We argue by induction on $n$. The cases $n = 0, 1, 2$ follow from Lemma 13.41.3. Assume $n > 2$. Suppose given a Postnikov system for the complex $X_{n - 1} \to X_{n - 2} \to \ldots \to X_0$. The only obstruction to extending this to a Postnikov system of length $n$ is that we have to find a morphism $X_ n \to Y_{n - 1}$ such that the composition $X_ n \to Y_{n - 1} \to X_{n - 1}$ is equal to the given map $X_ n \to X_{n - 1}$. Considering the distinguished triangle
and the associated long exact sequence coming from this and the functor $\mathop{\mathrm{Hom}}\nolimits (X_ n, -)$ (see Lemma 13.4.2) we find that it suffices to show that the composition $X_ n \to X_{n - 1} \to Y_{n - 2}$ is zero. Since we know that $X_ n \to X_{n - 1} \to X_{n - 2}$ is zero we can apply the distinguished triangle
to see that it suffices if $\mathop{\mathrm{Hom}}\nolimits (X_ n, Y_{n - 3}[-1]) = 0$. Arguing exactly as in the proof of Lemma 13.41.4 part (1) the reader easily sees this follows from the condition stated in the lemma.
The statement on isomorphisms follows from the existence of a map between the Postnikov systems extending the identity on the complex proven in Lemma 13.41.4 part (2) and Lemma 13.4.3 to show all the maps are isomorphisms. $\square$
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