Additive Kan maps and the normalized fibration criterion

Every simplicial abelian group is Kan. A horn in $X$ prescribes $x_i\in X_{n-1}$ for $i\mathrm{\ne }k$ with the compatibility $d_i{x}_j=d_{j-1}{x}_i$ for $i<j$, $i,j\mathrm{\ne }k$. Beginning with $u=0$, for $i=0,\dots ,k-1$ replace $u$ by $u+s_i(x_i-d_{i}u)$: the replacement fixes face $i$ because $d_i{s}_i=\mathrm{i}\mathrm{d}$, and it preserves all earlier faces because for $j<i$ the identity $d_j(x_i-d_{i}u)=d_{i-1}(x_j-d_{j}u)=0$ holds and $d_j{s}_i=s_{i-1}{d}_j$. Then for $i=n,n-1,\dots ,k+1$ replace $u$ by $u+s_{i-1}(x_i-d_{i}u)$, which fixes face $i$ since $d_i{s}_{i-1}=\mathrm{i}\mathrm{d}$ and preserves every already fixed face $j<k$ and $j>i$ by the same compatibility identities. The resulting $u$ fills every prescribed face, so $X$ is Kan.

Kan implies normalized surjectivity. Let $f$ be a Kan fibration and let $y\in N(Y{)}_n$, $n>0$. Use the zero horn ${\mathrm{\Lambda }}^n[n]\to X$ and target simplex $y\text{\hspace{0.17em}⁣}:\mathrm{\Delta }[n]\to Y$; its faces $d_{i}y$ for $i<n$ are zero, so this is a commutative horn square. A horn lift $x\in X_n$ satisfies $f(x)=y$ and $d_{i}x=0$ for $i<n$, hence $x\in N(X{)}_n$; thus $N(f{)}_n$ is surjective.

Termwise surjective additive maps are Kan. If $f\text{\hspace{0.17em}⁣}:X\to Y$ is termwise surjective and a horn in $Y$ is given, lift its target simplex $y$ to some $z\in X_n$, subtract the faces of $z$ from the prescribed horn to obtain a compatible horn in the kernel $\ker f$ (which is a simplicial abelian group, hence Kan by step 1.1), fill that horn by step 1.1, and add the filler to $z$. The result is a horn filler in $X$.

Boundary lifting from Kan plus quasi-isomorphism. Suppose $f$ is Kan and $N(f)$ is a quasi-isomorphism. Positive normalized degrees surject by step 1.2. In degree zero, given $y_0\in Y_0$ choose $x_0\in X_0$ with the same class in $H_0$ and write $y_0-f(x_0)=\mathrm{\partial }v$ for some $v\in N(Y{)}_1$; lifting $v$ to $N(X{)}_1$ by step 1.2 and correcting $x_0$ gives degree-zero surjectivity, so all normalized degrees are surjective and the Dold-Kan decomposition makes $f$ termwise surjective. The kernel $K$ of $f$ then has acyclic normalization by the exact sequence of normalized complexes, and an explicit boundary-filling argument applies: lift the target simplex, reduce to a boundary in $K$, fill faces $0,\dots ,n-1$ by successive degeneracy corrections as in step 1.1, and use acyclicity of $N(K)$ to correct the last normalized discrepancy. For $n=0$ termwise surjectivity suffices. Hence $f$ has boundary lifting.

Normalized surjectivity implies Kan. Assume $N(f{)}_n$ is surjective for all $n>0$, and let $D_n\subseteq Y_n$ be the set of simplices whose vertex component in ${\pi }_0(Y)=Y_0\mathrm{/}\mathrm{\partial }N(Y{)}_1$ lies in the image of ${\pi }_0(X)$; every vertex of a simplex has the same component, since successive vertices are joined by an edge whose difference is a boundary, so $D$ is a simplicial subgroup and a union of components, and $f$ maps into $D$. By the Dold-Kan decomposition of [F2], an element of $D_n$ lifts to $X_n$: write it in the summands $N(Y{)}_r$ ($r>0$), lift each coefficient by the hypothesis, and for the $r=0$ coefficient use that its component lies in the image of ${\pi }_0(X)$, choosing $x_0\in X_0$ and $v\in N(Y{)}_1$ with $y_0-f(x_0)=\mathrm{\partial }v$, lifting $v$ to $N(X{)}_1$ and correcting $x_0$. Hence $f\text{\hspace{0.17em}⁣}:X\to D$ is termwise surjective and is Kan by step 2.1. A horn square for $f$ with $n\ge 1$ has a nonempty horn, so its target simplex has a vertex and therefore lies in $D$; filling it over $D$ by step 2.1 fills it over $Y$.

Converse. Let $f$ have boundary lifting. Then it has horn lifting, and by the boundary-lifting criterion of [F3] it lifts every monomorphism, in particular the empty inclusions $\mathrm{\varnothing }\subset \mathrm{\Delta }[n]$ (a degreewise surjectivity statement) and $\mathrm{\varnothing }\subset Y$ (giving a section $s$ of $f$). Lifting the inclusion $X\times \mathrm{\partial }\mathrm{\Delta }[1]\subseteq X\times \mathrm{\Delta }[1]$ with endpoints ${\mathrm{i}\mathrm{d}}_X$ and $sf$ gives a homotopy ${\mathrm{i}\mathrm{d}}_X\simeq sf$, while $fs={\mathrm{i}\mathrm{d}}_Y$; the prism and free-additive homology argument of [F3] then shows that $N(f)$ is a quasi-isomorphism. Combining with step 2.2, boundary lifting is exactly Kan plus a quasi-isomorphism on normalized complexes, and the criterion applies to simplicial modules and to unital or nonunital simplicial algebras through their underlying additive groups.