Bocksteins Steenrod Squares and Cohomology Operations

The cochain $a$ in [F1] is a cocycle. Indeed, $i_{\ast }(\delta a)=\delta (i_{\ast }a)={\delta }^{2}b=0$; injectivity of $i_{\ast }$ gives $\delta a=0$.

Changing only the lift changes $a$ by a coboundary. If $b^{\mathrm{\prime }}$ is another lift of $c$, then $q_{\ast }(b^{\mathrm{\prime }}-b)=0$. Exactness gives a unique $h\in C^n(X;A)$ with $b^{\mathrm{\prime }}-b=i_{\ast }h$. If $a^{\mathrm{\prime }}$ is defined from $b^{\mathrm{\prime }}$, then $i_{\ast }(a^{\mathrm{\prime }}-a)=\delta (b^{\mathrm{\prime }}-b)=i_{\ast }(\delta h)$, hence $a^{\mathrm{\prime }}-a=\delta h$.

Changing the cocycle representative also changes $a$ by a coboundary. Write $c^{\mathrm{\prime }}=c+\delta u$. Use the lifting choice of [F2] to take $v\in C^{n-1}(X;B)$ with $q_{\ast }v=u$. For an arbitrary lift $b^{\mathrm{\prime }}$ of $c^{\mathrm{\prime }}$,

Steps 2.1 and 3.1 show that $[a^{\mathrm{\prime }}]=[a]$ for either permitted change, so $\beta [c]$ is well defined.

The Bockstein is natural in spaces. For $f\text{\hspace{0.17em}⁣}:X\to Y$, if $q_{\ast }b=c$ and $i_{\ast }a=\delta b$ on $Y$, then $q_{\ast }{f}^{\ast }b=f^{\ast }c$ and $i_{\ast }{f}^{\ast }a=\delta f^{\ast }b$. Therefore ${\beta }_X(f^{\ast }[c])=f^{\ast }{\beta }_Y[c]$; [F2] removes dependence on the displayed representatives.

The Bockstein is natural in the coefficient sequence. For a commutative morphism of short exact sequences with vertical maps $u\text{\hspace{0.17em}⁣}:A\to A^{\mathrm{\prime }}$, $v\text{\hspace{0.17em}⁣}:B\to B^{\mathrm{\prime }}$, and $w\text{\hspace{0.17em}⁣}:C\to C^{\mathrm{\prime }}$, the cochain $v_{\ast }b$ lifts $w_{\ast }c$, and $\delta (v_{\ast }b)=v_{\ast }{i}_{\ast }a=i_{\ast }^{\mathrm{\prime }}{u}_{\ast }a$. Hence ${\beta }^{\mathrm{\prime }}[w_{\ast }c]=u_{\ast }\beta [c]$.

The cone quotient comparison defines the signed suspension. Take a based CW complex $X$. If its chosen basepoint lies inside a positive-dimensional open cell, subdivide that one characteristic disk radially at the point and retain the same attaching maps for higher cells; this finite refinement makes the basepoint a vertex without changing the based space or choosing any new data. Form $CX=(X\times I)\mathrm{/}(X\times \{1\}\cup \{x_0\}\times I)$ and $\mathrm{\Sigma }X=CX\mathrm{/}X$. For each nonbasepoint $n$-cell of $X$, its product with the open height interval is an $(n+1)$-cell of $CX$; the height-zero cells form the cone base $X$, while the height-one face and basepoint track collapse to one vertex. The product characteristic maps have finite boundary-cell support, so their quotient map-out and CW weak-topology tests make $CX$ a CW complex with $X$ a closed subcomplex. A cellwise radial collar of this subcomplex, extended over the characteristic disks and assembled by the weak topology, gives an open neighborhood $V$ that strongly deformation retracts onto $X$. Since $V$ contains the whole fibre $X$ collapsed by $q:CX\to \mathrm{\Sigma }X$, it is saturated; $q(V)$ is open and its descended flow retracts onto the quotient vertex.

The coefficient connector anticommutes with the unsigned cone-pair connector. Take $q_{\ast }b=c$ on $X$, with $i_{\ast }a=\delta b$. Extend $c,b,a$ by zero on the singular simplices of $CX$ not lying in $X$, writing the extensions with bars. Then $d:=\delta \bar{b}-i_{\ast }\bar{a}$ is a relative $B$-cochain lifting the relative cocycle $\delta \bar{c}$, and

The signed suspension commutes with the Bockstein. For $x\in {\tilde{H}}^n(X;C)$,

Choose the canonical lifts $\tilde{\phi }$ and $\tilde{\psi }$ with values in $\mathbb{Z}\mathrm{/}{m}^2$. Because $\phi$ and $\psi$ are cocycles, there are uniquely determined $\mathbb{Z}\mathrm{/}m$-cochains $\eta$ and $\mu$ such that $\delta \tilde{\phi }=m\eta$ and $\delta \tilde{\psi }=m\mu$ in $\mathbb{Z}\mathrm{/}{m}^2$.

These cochains represent the two Bocksteins. By the defining lift-and-coboundary construction, $[\eta ]=\beta (x)$ and $[\mu ]=\beta (y)$.

Compute the Bockstein of the product. The cochain $\tilde{\phi }⌣\tilde{\psi }$ lifts $\phi ⌣\psi$, and [F2] gives

Divide by the injective copy of $\mathbb{Z}\mathrm{/}m$ and pass to cohomology. This yields the stated derivation identity. When $m=2$, $-1=1$ in the coefficient ring, so the parity sign is invisible.

Fix an explicit contraction on every standard diagonal carrier. The straight-line contraction of ${\mathrm{\Delta }}^n\times {\mathrm{\Delta }}^n$ to $(v_0,v_0)$ has the standard finite singular-prism chain homotopy $s_n$. Transport $s_n$ through the specified shuffle and Alexander--Whitney maps and add the specified homotopy from their composite to the identity. This gives a fixed map $h_n$ on $C_{\ast }({\mathrm{\Delta }}^n)\otimes C_{\ast }({\mathrm{\Delta }}^n)$ satisfying

Construct $D_i$ recursively. Let $W$ be the free ${\mathbb{F}}_2[C_2]$-resolution with one generator $e_i$ in each degree and $d{e}_i=(1+T)e_{i-1}$ for $i>0$. Put $D_0=\mathrm{AW}{\mathrm{\Delta }}_{\mathrm{\#}}$ as in [F1]. Suppose lexicographically that $D_i$ is known on lower-dimensional simplices and that $D_{i-1}$ is known. For the identity simplex ${\iota }_n$ set

The construction is natural and preserves subspaces. Postcomposition sends the formula for a simplex $\sigma$ to the formula for $f\sigma$, proving naturality. If the image of $\sigma$ lies in $A$, both tensor factors in step 2.1 lie in $C_{\ast }(A)$, proving the carrier assertion. This also includes degenerate singular simplices; none was quotiented out.

The same induction one degree higher proves coherent uniqueness. For two systems, subtract their recursive equations and suppose that $K_{i-1}$ is known while $K_i$ is already defined on every chain below the current dimension. On the identity simplex ${\iota }_n$ put $\omega =(D_i-D_i^{\mathrm{\prime }})({\iota }_n)+(1+T)K_{i-1}({\iota }_n)+K_i(d{\iota }_n)\mathrm{.}$ The recursion $d(D_i-D_i^{\mathrm{\prime }})+(D_i-D_i^{\mathrm{\prime }})d=(1+T)(D_{i-1}-D_{i-1}^{\mathrm{\prime }})$ for the two systems, the induction hypothesis for $K_{i-1}$, and the induction hypothesis for $K_i$ on the lower-dimensional chain $d{\iota }_n$ give $d\omega =0$: the two copies of $(1+T)d{K}_{i-1}({\iota }_n)$ cancel, while $(D_i-D_i^{\mathrm{\prime }})(d{\iota }_n)+(1+T)K_{i-1}(d{\iota }_n)$ and the explicit $d{K}_i(d{\iota }_n)$ agree by that second induction hypothesis. The element $\omega$ is therefore a cycle in the same standard carrier, and it has positive degree, so applying $h_n$ fills it and defines $K_i({\iota }_n)$; postcomposition extends it naturally. Taking the boundary of that defining filling gives exactly $D_i-D_i^{\mathrm{\prime }}=d{K}_i+K_{i}d+(1+T)K_{i-1}$.

If $i<0$, every cup product in the asserted formula has negative index, so [F1] makes both sides zero. Hence assume $i\ge 0$ and evaluate the left side on a chain $c$ of degree $p+q-i+1$. By the cochain-coboundary convention,

Substitute the higher-diagonal recurrence. Over ${\mathbb{F}}_2$ it gives

Expand the two terms. The tensor coboundary has no surviving signs over ${\mathbb{F}}_2$, so its first term is $(\delta a{⌣}_{i}b+a{⌣}_i\delta b)(c)$. Since $(a\otimes b)T=b\otimes a$, the second is $(a{⌣}_{i-1}b+b{⌣}_{i-1}a)(c)$. This proves the identity on every chain. The carrier property keeps every term relative when either input is relative. For $i=0$, both negative-index terms are zero and the formula reduces to the ordinary cup-product Leibniz identity.

If $k<0$ or $k>n$, [F1] makes $S{q}^k$ the zero homomorphism, so independence, additivity, and naturality are immediate. Hence assume $0\le k\le n$, so $j=n-k\ge 0$. The operation is additive. Expanding $(a+b){⌣}_j(a+b)$ leaves, besides the two individual squares, the cross term $a{⌣}_{j}b+b{⌣}_{j}a$. Since $a,b$ are cocycles, [F2] says

The class is independent of the coherently carried system. Pair $D_j-D_j^{\mathrm{\prime }}=d{K}_j+K_{j}d+(1+T)K_{j-1}$ with $a\otimes a$. The $d{K}_j$ term vanishes because $a\otimes a$ is a cocycle, the $K_{j}d$ term is the coboundary of $c\mapsto (a\otimes a)K_{j}c$, and the final term is zero because $(a\otimes a)T=a\otimes a$ and $2=0$.

The representing cochains are natural for spaces and pairs. For $f\text{\hspace{0.17em}⁣}:X\to Y$, naturality of $D_j$ makes $(f^{\ast }a\otimes f^{\ast }a)D_j^X=(a\otimes a)D_j^Y{f}_{\mathrm{\#}}$, so the representing cochains agree. For a map of pairs, [F4] makes the same equation descend to relative cochains.

The class is independent of its cocycle representative. If $a^{\mathrm{\prime }}=a+\delta h$, expansion and two applications of [F2] give

First prove $S{q}^0=\mathrm{id}$. Use the standard face-formula system of Medina--Mardones, Definition 7 and Theorem 10 (printed pages 8--9). On an $m$-simplex $s$ it is

Instability and the top square follow at the two definition endpoints. If $k>n$, [F2] declares $S{q}^{k}x=0$. If $k=n$, its representing cochain is $a{⌣}_{0}a$, which is the ordinary cup product by [F3]. Therefore $S{q}^{n}x=x⌣x$. The same argument uses the relative products when $x$ is relative.

Represent the cone-pair connector without a choice. Extend the cocycle $a$ from the cone base $X\subset CX$ to a cochain $b$ on $CX$ by setting it to zero on every singular simplex not lying in $X$. Then $c:=\delta b$ vanishes on chains in $X$ and so is a relative cocycle in $C^{n+1}(CX,X;{\mathbb{F}}_2)$. By [F5], $[c]=\mathrm{\partial }[a]$. This extension is a specified function, not an application of AC.

The connector commutes with every square. For $0\le k\le n$, set $j=n-k$ and define

The cone-pair connector is the reduced suspension after the quotient comparison. Take a based CW complex. If its basepoint lies inside a positive-dimensional open cell, radially subdivide that characteristic disk there and keep the higher attaching maps; this finite refinement makes it a vertex without changing the based space. For every nonbasepoint $n$-cell of $X$, its product with the open height interval in [F6] gives an $(n+1)$-cell of $CX$; the height-zero cells form the copy of $X$, while $X\times \{1\}$ and the basepoint track collapse to one vertex. The product characteristic disks supply the attaching maps, and each has finite boundary-cell support because its $X$-cell does. Their quotient map-out test and the CW weak topology give the cone its CW structure, with $X$ a closed subcomplex. The cellwise radial collar of this subcomplex gives an open neighborhood $V$ that strongly deformation retracts onto $X$: extend the collar and its radial flow over each characteristic disk, and assemble the compatible extensions using the CW weak topology. Since $X\subset V$ is the entire collapsed fibre of $q:CX\to \mathrm{\Sigma }X$, $V$ is saturated; hence $q(V)$ is open and the flow descends to a retraction onto the quotient vertex.

Record the diagonal on the mod-two $C_2$ resolution. Let $W$ have one free-orbit generator $e_i$ in every degree $i\ge 0$, with $d{e}_i=(1+T)e_{i-1}$ and $e_{-1}=0$. Define

Fix an explicit contraction of every common fourfold model carrier. For a standard simplex ${\mathrm{\Delta }}^m$, let $P_m$ be the prism from its affine contraction to the first vertex. By [F4], $d{P}_m+P_{m}d=1-j_m{p}_m$, where $p_m$ collapses to a point and $j_m$ includes that vertex. On the unnormalized point complex, whose degree-$n$ generator is $e_n$, put $a(e_n)=e_{n+1}$ for odd $n$ and $a(e_n)=0$ for even $n$. Directly, $da+ad=1-\eta ϵ$. Hence

Assemble the two displayed families into equivariant maps. Define $\mathcal{L}(e_i\otimes z)=L_i(z)$. This is the composite obtained by shuffling $z$ to $X\times Y$, applying the equivariant higher diagonal there, and applying $A$ to its two outputs. Define $\mathcal{R}$ by first applying $\rho$, then applying the two higher-diagonal systems and finally regrouping with $\tau$. The factor $T^r$ in $R_i$ is precisely the twist in $\rho$. Naturality and the chain-map identities in [F1]--[F3], together with step 1.1, show that both are $C_2$-equivariant chain maps