# Proof model category chain complexes

In the model category of chain complexes one defines a coﬁbration to be a chain map $M\to N$ such that for each $k>0$ the map $f_k:M_k\to N_k$ is a monomorphism with a projective $R$-module as its cokernel, and a ﬁbration if for each $k > 1$ the map $f_k : M_k → N_k$ is an epimorphism. Weak equivalences are quasi isomorphisms.

Now I want to show that cofibrations have the LLP with respect to trivial fibrations. So look at a diagram

$\require{AMScd} \begin{CD} A @>{g}>> X;\\ @V{i}VV @VV{p}V \\ B @>{h}>> Y; \end{CD}$

where $i$ is a cofibration and $p$ is a trivial fibration. In particular $p$ is surjective in all degrees.

Dwyer and Spalinsky define the map degree by degree. In zeroth degree: By assumption on $i$ the image splits into $A_0\oplus P_0$ where $P_0$ is projective. Now define the lifting $f$ on $P_0$ to be any lifting of the map $p$ and define the lifting to be $g_0$ on $A_0$.

My question: Why does the relation $g_0=f_0\circ i_0$ hold? This makes sense if the composition of $i_0$ with the isomorphism of $B_0\cong A_0\oplus P_0$ is the inclusion, but why does this hold?

That's just what you mean by the image splitting. $i_0$ is a split monomorphism, so there's a biproduct structure on $B_0$, canonical up to a choice of splitting, whose inclusion at $A_0$ is $i_0$ and whose projection to $P_0$ is a chosen cokernel map. You should prove this if it doesn't strike you as clear.
• Thanks for your answer. So this seems to be a pure category theoretical question. Can you be more explicit why $i_0$ is the inclusion at $A_0$ of $B_0$? – userZ617 Aug 8 '16 at 22:57
• I tried to show that for any map $t:A_0\to C_0$ there is a unique map $\psi:B_0\to A_0\oplus P_0\to C_0$ such that $t=\psi\circ i$. Is this useful and how do I choose the map? – userZ617 Aug 8 '16 at 23:37