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A map that is both a model fibration and a quasi-isomorphism. The homology isomorphism together with degree-one surjectivity forces surjectivity in degree zero. Equivalently it has the right lifting property against the generating cofibrations for nonnegative chain complexes.

Use homological grading $d_n:C_n\to C_{n-1}$, with zero terms below degree zero. In the projective model structure on nonnegative chain complexes, the classes are:

Thus a cofibrant complex has a projective module in each degree, and every object is fibrant. An projective acyclic fibration is equivalently a degreewise surjective quasi-isomorphism, including degree zero. To see the degree-zero assertion, lift a degree-zero homology class using the homology isomorphism, then lift the boundary discrepancy using surjectivity in degree one. Conversely a degreewise surjective quasi-isomorphism has the requisite positive-degree surjections.

Let $S^n(R)$ be the sphere chain complex with $R$ in degree $n$ and zero elsewhere. Let $D^n(R)$, for $n\ge 1$, be the disk chain complex with $R$ in degrees $n,n-1$ and identity differential. The generating cofibrations for nonnegative chain complexes areEach is a model cofibration: its degreewise cokernel is a copy of $R$ in one degree. The extra degree-zero generator must not be omitted.

Here is the lifting characterization of an acyclic chain-complex fibration. A right lifting property for $S^{n-1}(R)\to D^n(R)$ says that whenever $x\in Z_{n-1}X$ and $y\in Y_n$ satisfy $f(x)=dy$, there is $z\in X_n$ with $dz=x$ and $f(z)=y$. The generator $0\to S^0(R)$ says that $f_0$ is surjective. For a degreewise surjective quasi-isomorphism, its kernel is acyclic by the long exact sequence in homology. Choose a preliminary lift $z_0$ of $y$. Then $d{z}_0-x$ is a cycle in the kernel; subtract an element of the kernel whose boundary is this cycle. This gives the required $z$.

Conversely the lifting conditions make the kernel acyclic: take $y=0$ and any kernel cycle $x$. They also imply degreewise surjectivity inductively. Having surjectivity below degree $n$, lift $dy$ to an element of $X_{n-1}$, correct its boundary within the acyclic kernel to make it a cycle, and apply the lifting condition to lift $y$. For $n=1$ the chosen degree-zero element is already a cycle. A degreewise surjective map with acyclic kernel is a quasi-isomorphism by the long exact sequence in homology. Hence $I$-injectives are exactly acyclic fibrations.

An object $A$ is sequentially small if for every sequential diagram $Z_0\to Z_1\to \cdots$, the natural mapis bijective. Surjectivity says that a map out of $A$ factors through a finite stage; injectivity says that two such maps agreeing in the colimit agree at a later finite stage. The relative version restricts to the stated class of sequential diagrams. The domains in $I$ have this property: maps from $S^{n-1}(R)$ are cycles in one degree, and filtered colimits commute with these finite equations. Their sequential smallness is also permitted by the hint.

Apply the small object argument to the map $a:X\to Y$. Put $Z_0=X$, $p_0=a$. Given $p_k:Z_k\to Y$, take the set of all commutative squares with a generator $i:A_i\to B_i$ on the left and $p_k$ on the right. Form the pushout of the coproduct of all these $i$ along their maps into $Z_k$, obtaining $Z_{k+1}$. Each bottom map $B_i\to Y$ induces the compatible map $p_{k+1}:Z_{k+1}\to Y$. Repeat for all $k\ge 0$, and set $Z_{\infty }={\mathrm{colim}}_k{Z}_k$. This givesConcretely, attaching $S^{n-1}(R)\to D^n(R)$ adds a free generator in degree $n$ whose boundary is the chosen existing cycle; attaching $0\to S^0(R)$ adds a free degree-zero generator. Thus $j$ is an injection, and its cokernel in each degree is free on the newly attached generators. It is therefore a model cofibration; this also illustrates the relative cell complex description of the left factor.

For a lifting square into $Z_{\infty }\to Y$, sequential smallness of its domain $A_i$ factors the top map through some $Z_k$. The commutativity is already an equality in $Y$, so this is one of the squares attached at stage $k+1$. Its new cell supplies a lift $B_i\to Z_{k+1}\to Z_{\infty }$. Hence $p$ has the right lifting property with respect to every element of $I$, and is an projective acyclic fibration. Every map has the required cofibration–acyclic-fibration factorization. Only the domains need sequential smallness; there is no requirement that $X$, $Y$ or the coproduct of all cells be small.