ex-a-canonical-truncation-triangle.md

The kernel and cokernel of multiplication by two are $2R$ and $R\mathrm{/}2R$. Set $A=(2R)[0]$. The quotient complex $V=X\mathrm{/}A$ has $V^0=R\mathrm{/}2R,V^1=R$ with differential $\bar{r}\mapsto 2r$. The map $v:V\to (R\mathrm{/}2R)[-1]$ is quotient in degree one and zero in degree zero. Its kernel is the identity complex on $2R$ after identifying the degree-zero term with $2R$, so $v$ is a quasi-isomorphism.

Let $C=\mathrm{Cone}(i)$, with $C^k=X^k\oplus A^{k+1}$ and differential $(x,a)\mapsto (d_{X}x+i(a),-d_{A}a)$. The cone-to-quotient map $e:C\to V$ sends $(x,a)$ to the class of $x$; its kernel is the contractible identity cone on $A$. Thus $e$ is a quasi-isomorphism. Let $p:C\to A[1]$ be $(x,a)\mapsto a$. Then $\delta$ is the roof $(R\mathrm{/}2R)[-1]\stackrel{ve}{\leftarrow }C\stackrel{p}{\to }A[1]$. The cone triangle transported through $ve$ has first arrow $i$ and second arrow $q=v(X\to V)$, proving every displayed arrow with the stated signs.