= Solution
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:
* <Model weak equivalences> are <quasi-isomorphisms>, inducing isomorphisms on every $H_n$.
* <Model fibrations> are maps surjective in every strictly positive degree $n\geq1$; surjectivity in degree zero is not part of this definition.
* <Model cofibrations> are degreewise injections with projective cokernel in every degree. The underlying module injections then split degreewise, although these splittings need not commute with the differential.
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 $n\geq1$, be the <disk chain complex> with $R$ in degrees $n,n-1$ and identity differential. The <generating cofibrations for nonnegative chain complexes> are
$$
\boxed{I=\{0\to S^0(R)\}\ \cup\ \{S^{n-1}(R)\hookrightarrow D^n(R):n\geq1\}.}
$$
Each 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 $dz_0-x$ 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 \b[$I$-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 map
$$
\boxed{\operatorname*{colim}_k\operatorname{Hom}(A,Z_k)\ \longrightarrow\ \operatorname{Hom}\left(A,\operatorname*{colim}_k Z_k\right)}
$$
is 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 $k\geq0$, and set $Z_\infty=\operatorname*{colim}_k Z_k$. This gives
$$
\boxed{X\xrightarrow{j}Z_\infty\xrightarrow{p}Y,\qquad pj=a.}
$$
Concretely, 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>. \b[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.
Back to article page
