Eq. (13) — Exact distance with a one-complex

For every finite-length based binary chain complex $A$ over $F_2 = Z/2Z$, every pair of natural numbers $r,c$ (including zero), every $F_2$-linear map $P:F_2^c → F_2^r$, and every natural number $j$, the declaration asserts the exact equality in $WithTop(N)=N ∪ {∞}$:

where $K$ is the one-complex constructed from $P$. These are all the quantified arguments, and there are no further hypotheses on $A,P,r,c$, or $j$. Concretely, $A$ consists of dimensions $dim_A(n) ∈ N$, binary word spaces $A_n=(Fin(dim_A(n)) → F_2)$, and boundary maps indexed by their domain degree: $∂^A_0:A_0 → (Empty → F_2)$, necessarily the zero map to the one-element zero vector space, and $∂^A_{n+1}:A_{n+1} → A_n$ for $n ≥ 0$. They satisfy $∂^A_n ∘ ∂^A_{n+1}=0$ for every $n$. The structure also has a natural-number length $L_A$ such that $dim_A(n)=0$ whenever $L_A<n$, but $L_A$ need not be minimal and dimensions at or below it may also vanish. For any such complex $X$, its degree-$n$ distance is the infimum in $N ∪ {∞}$ of the Hamming weights of exactly those $x ∈ X_n$ for which $∂^X_nx=0$ and $x$ is not in the range of $∂^X_{n+1}$. This distance is $∞$ when that candidate set is empty and otherwise is the least occurring finite Hamming weight. It is never zero, because a word of weight zero is the zero word and zero belongs to every linear-map range. At degree zero, $∂^X_0=0$, so every word is a cycle and the candidates are precisely the words outside the range of $∂^X_1$. The one-complex $K$ has $K_0=F_2^r$, $K_1=F_2^c$, and the zero vector space in every degree $n ≥ 2$; its boundaries are $∂^K_0=0$, $∂^K_1=P$, and $∂^K_n=0$ for $n ≥ 2$, and its recorded length is $1$, even if one or both displayed spaces vanish. Consequently,

so $d_1(K)=∞$ exactly when $P$ is injective, while

so $d_0(K)=∞$ exactly when $P$ is surjective. The quantity $d_{<j}(A)$ is $∞$ when $j=0$ and is $d_k(A)$ when $j=k+1$; it does not introduce an actual negative-degree chain group. The tensor degree-$n$ coordinate type is the dependent disjoint union of $Fin(dim_A(p)) × Fin(dim_K(q))$ over all ordered pairs of nonnegative degrees $(p,q)$ satisfying exactly $p+q=n$, with each pair retained as a tag. Since $K_q$ is zero for $q ≥ 2$, the only potentially nonempty components are $(p,q)=(n,0)$, using coordinates of $A_n$ and $F_2^r$, and, when $n ≥ 1$, $(p,q)=(n-1,1)$, using coordinates of $A_{n-1}$ and $F_2^c$; either component can still be empty when one of its dimensions is zero. Write $T_n$ for the binary word space on this tensor coordinate type. The tensor boundary indexed by its domain is $Δ_0:T_0 → (Empty → F_2)$, defined to be zero, and $Δ_{n+1}:T_{n+1} → T_n$ for $n ≥ 0$. At a target coordinate tagged by $(p,q)$ with $p+q=n$, basis coordinates $a$ of $A_p$ and $b$ of $K_q$, and source word $e ∈ T_{n+1}$, its value is exactly

The first source coordinate is tagged by $(p+1,q)$, the second by $(p,q+1)$, and the two contributions are added in $F_2$, with no sign factor. For the one-complex, the second contribution uses $P$ exactly when $q=0$ and is zero when $q ≥ 1$. The imported declarations prove $Δ_n ∘ Δ_{n+1}=0$ for every $n$, including the endpoint involving $Δ_0$. The left-hand side $D_j(A ⊗ K)$ is the infimum of the Hamming weights of exactly those $e ∈ T_j$ satisfying $Δ_je=0$ and not belonging to the range of $Δ_{j+1}$. It is $∞$ if this candidate set is empty, including when $T_j$ is the zero vector space, and otherwise is a positive finite minimum. At $j=0$, the only degree pair is $(0,0)$, $Δ_0=0$, and every tensor word is a cycle. Also $d_{<0}(A)=∞$; because $d_1(K)$ is never zero, the first product on the right is $∞$, so the asserted equality specializes exactly to

At a positive degree $j=k+1$, it specializes exactly to

All products and minima are taken in $N ∪ {∞}$. Because none of the distance factors is zero, a displayed product is finite exactly when both factors are finite and is $∞$ when either factor is $∞$; the binary minimum is finite if at least one product is finite and is $∞$ exactly when both products are $∞$. Thus injectivity of $P$ makes the $d_1(K)$ product infinite, surjectivity makes the $d_0(K)$ product infinite, and bijectivity makes both products infinite, forcing the asserted tensor distance to be $∞$. If $r=0$, then $P$ is automatically surjective and $d_0(K)=∞$; if $c=0$, then $P$ is automatically injective and $d_1(K)=∞$. The theorem imposes no rank condition, no positivity condition on $r$ or $c$, no nontrivial-homology or finite-distance condition, and no restriction of $j$ by the recorded length of $A$. It asserts equality, not merely either inequality, and its right-hand side is a binary minimum rather than a minimum over an additional index set.