Normalized simplicial chains, prism homotopies and the abelian trivial-fibration criterion

The normalization projection. Define $p_{n,0}=\mathrm{i}\mathrm{d}$ and $p_{n,i}=(1-s_{i-1}{d}_{i-1})\cdots (1-s_0{d}_0)$ for $1\le i\le n$, and $p_n=p_{n,n}$. Applying the factors successively kills $d_0,\dots ,d_{n-1}$: if $d_{j}x=0$ for $j<i$ then $d_j(1-s_i{d}_i)x=d_{j}x-d_j{s}_i{d}_{i}x=0$ for $j<i$ by the identities of [F1], while $d_i(1-s_i{d}_i)x=d_{i}x-d_i{s}_i{d}_{i}x=0$. Hence the image of $p_n$ lies in $N(M{)}_n$, and $p_n$ is the identity on $N(M{)}_n$ because each factor acts as the identity there.

Prism homotopies. Let $H\text{\hspace{0.17em}⁣}:X\times \mathrm{\Delta }[1]\to Y$ be a simplicial homotopy from $f$ to $g$ of simplicial sets. The prism maps $h_i(x)=H_{n+1}(s_{i}x,(0,\dots ,0,1,\dots ,1)),$ with $i+1$ zeros, for $x\in X_n$ and $0\le i\le n$, induce a chain homotopy ${\sum }_i(-1{)}^i{h}_i$ between the induced maps on free abelian (or free $R$-module) chains: expanding the boundary of ${\sum }_i(-1{)}^i{h}_i$, the internal face terms cancel in adjacent prism terms by the simplicial identities, and the surviving endpoint faces are exactly $g_{\mathrm{\#}}-f_{\mathrm{\#}}$. Consequently a simplicial homotopy equivalence of underlying simplicial sets induces a chain homotopy equivalence, hence a quasi-isomorphism, on free chains.

The chain homotopy, with a telescoping verification. For each $r\ge 0$ define $q_n^r=p_{n,\min (r,n)}$ on the Moore complex, so $q^0=1$. Inductively $q^r$ is a chain map and its degree-$n$ image has $d_i=0$ for $i<\min (r,n)$. Put $h_n^r=(-1{)}^r{s}_r{q}_n^r$ when $n\ge r$ and $h_n^r=0$ otherwise. For $n>r$, the simplicial identities and the vanished first $r$ faces give $\mathrm{\partial }{s}_r{q}_n^r=-s_r{\sum }_{j>r}(-1{)}^j{d}_j{q}_n^r$. Since $q^r$ is a chain map, adding $h_{n-1}^r\mathrm{\partial }=(-1{)}^r{s}_r\mathrm{\partial }{q}_n^r$ leaves exactly $s_r{d}_r{q}_n^r=q_n^r-q_n^{r+1}$. For $n=r$, the only possibly nonzero two faces of $s_r{q}_r^r$ cancel, and the previous homotopy term is zero; for $n<r$ all terms are zero. Thus $\mathrm{\partial }{h}^r+h^r\mathrm{\partial }=q^r-q^{r+1}$ in all degrees. This also proves that $q^{r+1}$ is a chain map, completing the induction from $q^0$. In a fixed degree the sequence stabilizes at $p_n$, so summing these homotopies gives $H_n={\sum }_{r=0}^n(-1{)}^r{s}_r{p}_{n,r}$ and $\mathrm{\partial }H+H\mathrm{\partial }=1-\iota p$. The projection $p$ is a natural chain map into $N(M)$, is the identity there by step 1.1, and the identity proves the claimed natural chain homotopy equivalence.

From set homotopy equivalence to additive homology. Write $C(M)$ for the chain complex of the free simplicial abelian group $\mathbb{Z}[M]$, and let $e:C(M)\to s(M)$ send $[a]$ to $a$. If the underlying simplicial map of $f:M\to M^{\mathrm{\prime }}$ is a homotopy equivalence, step 1.2 shows that $C(f)$ is a homology isomorphism. Let $x\in N(M{)}_n$ be a normalized cycle. Every face of $x$ is zero (for $n=0$ there are no faces), so $c_x=[x]-[0]$ is a normalized free cycle and $e(c_x)=x$. For injectivity, suppose $f(x)$ is an additive boundary. By step 2.1 choose $y\in N(M^{\mathrm{\prime }}{)}_{n+1}$ with $\mathrm{\partial }y=f(x)$. Put $w=(-1{)}^{n+1}y$, so $d_{n+1}w=f(x)$ and $d_{i}w=0$ for $i<n+1$. Then the free chain $(-1{)}^{n+1}([w]-[0])$ has boundary exactly $[f(x)]-[0]$. Hence $C(f)(c_x)$ is a free boundary, so $c_x$ is a free boundary by injectivity on free homology; evaluation makes $x$ an additive boundary. For surjectivity, start with a normalized cycle $z\in N(M^{\mathrm{\prime }}{)}_n$. The free cycle $[z]-[0]$ has a homology preimage represented by some free cycle $c\in C(M{)}_n$; no claim is made that $c$ is one basis difference. Since $C(f)(c)-([z]-[0])$ is a free boundary, evaluation shows that $f(e(c))-z$ is an additive boundary. Project $e(c)$ into $N(M)$ using step 2.1 if necessary. This proves surjectivity. The argument includes arbitrary additive degree-zero cycles, and its bounding-chain formula uses the normalized last face only after correcting the sign.

Exactness of normalization. If $f\text{\hspace{0.17em}⁣}:M\to M^{\mathrm{\prime }}$ is degreewise surjective, then $N(f)\text{\hspace{0.17em}⁣}:N(M)\to N(M^{\mathrm{\prime }})$ is surjective: given a normalized $y\in N(M^{\mathrm{\prime }}{)}_n$, choose $x\in s(M{)}_n$ with $f(x)=y$; naturality of $p_n$ gives $f(p_{n}x)=p_{n}f(x)=p_{n}y=y$ because $y$ is normalized and $p_n$ is the identity on normalized elements. Hence $N$ is exact, since it is a functor that preserves kernels and turns degreewise epimorphisms into epimorphisms, so it preserves short exact sequences of simplicial abelian groups in each degree.

The trivial-fibration criterion. Let $f\text{\hspace{0.17em}⁣}:M\to M^{\mathrm{\prime }}$ be termwise surjective and a quasi-isomorphism. Its kernel $K=\ker f$ is acyclic by the long exact homology sequence of the degreewise short exact sequence of complexes (cycle lifts and boundary lifts give its elementary proof), and let a boundary-lifting problem with target simplex $v\in M_n^{\mathrm{\prime }}$ and prescribed faces $x_i\in M_{n-1}$, satisfying $f(x_i)=d_{i}v$ and $d_i{x}_j=d_{j-1}{x}_i$ for $i<j$, be given. In degree zero, choose a lift directly using termwise surjectivity. For $n\ge 1$, choose a lift $u_0\in M_n$ of the prescribed target simplex and replace it first by $u=u_0+s_0(x_0-d_0{u}_0)$ and $u\leftarrow u+s_r(x_r-d_{r}u)$ for $r=1,\dots ,n-1$, using the simplicial identities of [F1] to preserve the faces already filled and to fill the $r$-th face; the remaining discrepancy $z=x_n-d_{n}u\in K_{n-1}$ has all faces zero by construction, hence is a normalized cycle. Since $K$ is acyclic and $N(K)\to s(K)$ is a chain homotopy equivalence by step 2.1, the normalized cycle $z$ is a boundary already in $N(K)$: there is $w\in N(K{)}_n$ with $(-1{)}^n{d}_{n}w=z$, and then $u+(-1{)}^{n}w$ fills the last face while leaving the previously filled faces unchanged. This supplies every boundary lift, so $f$ is a trivial Kan fibration. Every step is an explicit formula, so no choice principle is used.