**Proof.**
We first prove (1) by induction on $j$. For the base case $j = 0$ there is nothing to prove as $Y'_0 \to X'_0$ is an isomorphism. Say the result holds for $j - 1$. We consider the distinguished triangle

\[ Y'_ j \to X'_ j \to Y'_{j - 1} \to Y'_ j[1] \]

The long exact sequence of Lemma 13.4.2 gives an exact sequence

\[ \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], Y'_{j - 1}[-1]) \to \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], Y'_ j) \to \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], X'_ j) \]

From the induction hypothesis and (13.40.4.1) we conclude the outer groups are zero and we win.

Proof of (2). For $n = 1$ the existence of morphisms has been established in Lemma 13.40.3. For $n > 1$ by induction, we may assume given the map of Postnikov systems of length $n - 1$. The problem is that we do not know whether the diagram

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

is commutative. Denote $\alpha : X_ n \to Y'_{n - 1}$ the difference. Then we do know that the composition of $\alpha $ with $Y'_{n - 1} \to X'_{n - 1}$ is zero (because of what it means to be a map of Postnikov systems of length $n - 1$). By the distinguished triangle $Y'_{n - 1} \to X'_{n - 1} \to Y'_{n - 2} \to Y'_{n - 1}[1]$, this means that $\alpha $ is the composition of $Y'_{n - 2}[-1] \to Y'_{n - 1}$ with a map $\alpha ' : X_ n \to Y'_{n - 2}[-1]$. Then (13.40.4.1) guarantees $\alpha '$ is zero by part (1) of the lemma. Thus $\alpha $ is zero. To finish the proof of existence, the commutativity guarantees we can choose the dotted arrow fitting into the diagram

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

by TR3.
$\square$

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