Splitting a bounded complex by vanishing higher Ext

First let $A\stackrel{u}{\to }B\stackrel{v}{\to }C\stackrel{w}{\to }A[1]$ be distinguished with $w=0$. Hom exactness supplies $s:C\to B$ with $vs=1_C$. The split triangle maps to this triangle by $(1_A,(u,s),1_C)$: the middle square uses $vu=0,vs=1$, and the last square uses $w=0$. Two isomorphism components force $(u,s)$ to be an isomorphism. Conversely such a split-triangle isomorphism forces $w=0$. This includes zero vertices.

Choose $a\le b$ bounding the cohomology. For empty cohomological support the object is zero, and for a single degree the canonical truncation maps identify $X$ with $H^a(X)[-a]$. If $a<b$, use the triangle ${\tau }^{\le b-1}X\to X\to H^b(X)[-b]\to ({\tau }^{\le b-1}X)[1]$. Induction on $b-a$ identifies its head with ${⨁}_{a\le i<b}{H}^i(X)[-i]$.

The connecting map lies in the finite direct sum of groups ${\mathrm{Hom}}_D(H^{b}X[-b],H^{i}X[-i+1])={\mathrm{Ext}}^{b-i+1}(H^{b}X,H^{i}X)$. Since $i<b$, every exponent is at least two, so each group vanishes by hypothesis. The splitting criterion in step 1.1 completes the induction. Its section $s$ was chosen and need not be unique.

Under the additional Yoneda hypotheses, any positive-degree Ext class is a Yoneda extension class. For $p>2$, break its exact extension at the image after the last two arrows toward its quotient endpoint. It becomes a splice of a two-extension and a $(p-2)$-extension. If all two-extension groups vanish, composition compatibility makes the splice zero. Thus all Ext groups of degree at least two vanish for all pairs, and step 2.1 applies.