A zero Yoneda splice factors through the middle object

Call a filling of the displayed two-extension an object $T$ with an exact sequence $0\to B\stackrel{b}{\to }T\stackrel{t}{\to }X\to 0$ and a map $r:T\to D$ such that $rb=jp$ and $qr=t$, for which $0\to A\stackrel{bi}{\to }T\stackrel{r}{\to }D\to 0$ is exact. A lift $G$ in the statement is exactly such a filling: its pushout along $p$ is $T\mathrm{/}A$, and an endpoint-fixed equivalence to $F$ is an isomorphism by [F2]. Conversely the quotient map $r$ identifies that pushout with $F$. Thus it suffices to prove that existence of a filling is invariant under each direction of an endpoint-fixed chain map of two-extensions.

Consider any such chain map with middle maps $u:B\to B^{\mathrm{\prime }}$ and $v:D\to D^{\mathrm{\prime }}$. Factoring the differentials through their images gives $c:C\to C^{\mathrm{\prime }}$ satisfying $ui=i^{\mathrm{\prime }}$, $p^{\mathrm{\prime }}u=cp$, $vj=j^{\mathrm{\prime }}c$, and $q^{\mathrm{\prime }}v=q$. The natural maps $B\to B^{\mathrm{\prime }}{\times }_{C^{\mathrm{\prime }}}C$ and $C^{\mathrm{\prime }}{⨿}_{C}D\to D^{\mathrm{\prime }}$ induce identities on the endpoints of the respective short exact sequences. They are isomorphisms by [F2]. In particular $B=B^{\mathrm{\prime }}{\times }_{C^{\mathrm{\prime }}}C$ and $F^{\mathrm{\prime }}=c_{\ast }F$, up to these specified isomorphisms.

If the unprimed two-extension has a filling $T$, push $0\to B\to T\to X\to 0$ out along $u$ to obtain $0\to B^{\mathrm{\prime }}\to T^{\mathrm{\prime }}\to X\to 0$. The maps $T\stackrel{r}{\to }D\stackrel{v}{\to }{D}^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}\stackrel{p^{\mathrm{\prime }}}{\to }{C}^{\mathrm{\prime }}\stackrel{j^{\mathrm{\prime }}}{\to }{D}^{\mathrm{\prime }}$ agree on $B$, so induce $r^{\mathrm{\prime }}:T^{\mathrm{\prime }}\to D^{\mathrm{\prime }}$. Quotienting the pushout by the common copy of $A$ gives $T^{\mathrm{\prime }}\mathrm{/}A\cong C^{\mathrm{\prime }}{⨿}_C(T\mathrm{/}A)\cong C^{\mathrm{\prime }}{⨿}_{C}D\cong D^{\mathrm{\prime }}$. These isomorphisms follow directly from the universal property of a cokernel and a pushout: a map out of either side is precisely a compatible pair of maps annihilating $A$. Consequently $\ker r^{\mathrm{\prime }}=A$ and $r^{\mathrm{\prime }}$ is epic. This is a filling of the primed extension.

If instead the primed extension has a filling $T^{\mathrm{\prime }}$, form $T=T^{\mathrm{\prime }}{\times }_{D^{\mathrm{\prime }}}D$ using its quotient $r^{\mathrm{\prime }}:T^{\mathrm{\prime }}\twoheadrightarrow D^{\mathrm{\prime }}$ and $v$. The projection $r:T\to D$ is epic with kernel $A$ by pullback exactness. Set $t=qr$, which is epic. Its kernel is $T^{\mathrm{\prime }}{\times }_{D^{\mathrm{\prime }}}C$, where $C\to D^{\mathrm{\prime }}$ is $j^{\mathrm{\prime }}c$. Since the inverse image of $C^{\mathrm{\prime }}\subset D^{\mathrm{\prime }}$ in $T^{\mathrm{\prime }}$ is $B^{\mathrm{\prime }}$, this kernel is $B^{\mathrm{\prime }}{\times }_{C^{\mathrm{\prime }}}C\cong B$ by step 1.2. The induced $B\to T$ satisfies $rb=jp$ and identifies its copy of $A$ with $\ker r$. Thus $T$ is a filling of the unprimed extension. Every object and morphism used here is a finite pullback or pushout; the argument uses neither resolutions nor a choice principle.

The zero representative $Z$ has filling $A\oplus X$, with its two evident split short exact sequences. If the given splice is zero, [F1] supplies an actual finite zigzag from it to $Z$. Traverse this zigzag backwards starting with that filling, applying step 2.1 when the traversal follows a chain map and step 2.2 when it opposes one. Finite induction produces a filling of the given splice, hence the required $G$. This explicitly supplies the factorization data through every switch, including switches whose intermediate kernels change.

Conversely let a filling $T$ be given. Form $H=B{\times }_{C}B$, with first projection $h:H\to B$ and second projection $s:H\to B$. The sequence $0\to A\stackrel{(i,0)}{\to }H\stackrel{s}{\to }B\to 0$ is split by the diagonal and has a retraction $\rho :H\to A$ specified by $i\rho =h-s$. Splice this split extension with $0\to B\stackrel{b}{\to }T\to X\to 0$ to obtain $L:0\to A\to H\stackrel{bs}{\to }T\to X\to 0$. There is an endpoint-fixed chain map $L\to EF$, with middle maps $h$ and $r$, since $jph=jps=rbs$. There is also an endpoint-fixed chain map $L\to Z$, with middle maps $\rho$ and $t$; the differential square commutes because $tbs=0$. Hence $EF\leftarrow L\to Z$ is a concrete finite zigzag, proving the splice is zero.