Finite duality and failure of opposite density

Finite-dimensional spaces test an arbitrary vector space in two different directions. Maps from them determine every linear map between the spaces being tested. Maps to them instead produce a completion: the algebraic double dual. Reversing arrows therefore changes the density question, even though finite-dimensional duality itself is an equivalence.

Let $k$ be any field, including a finite field. Write $\mathsf V=\mathsf{Vect}_k$ for vector spaces in a fixed universe, and choose a small full skeleton $\mathsf F$ of its finite-dimensional spaces. We use algebraic duals $V^*=\operatorname{Hom}_k(V,k)$, with no continuity requirement. Basis extension and extension of linear functionals use the usual axiom of choice. The prerequisite is Dense probes and reconstruction from colimits, particularly its enlargement lemma, finite free-probe result, and strict-quotient density criterion. Ordinary module kernels, cokernels and strictness are retained from Coimages, images and composition of quotients.

1. Finite duality and the canonical completion

The evaluation map is

\begin{gathered} \delta_V:V\longrightarrow V^{**},\\ \delta_V(v)(\lambda)=\lambda(v). \end{gathered} \tag{1.1}

It is injective: extend a nonzero vector to a basis and use its coordinate functional. For $f:V\to U$, evaluation gives $f^{**}\delta_V=\delta_U f$. If $V$ is finite-dimensional, a basis and its dual basis show that $\delta_V$ is also surjective. Consequently

D:\mathsf F^{\rm op}\longrightarrow\mathsf F, \qquad W\longmapsto W^* \tag{1.2}

is an equivalence. Applying duality twice gives a functor naturally isomorphic to the identity by (1.1). On a skeleton we choose the corresponding representatives; all statements are invariant under these choices.

For arbitrary $V$, let $\mathsf Q_V$ have objects $q:V\to W$, with $W\in\mathsf F$. A morphism from $q$ to $q'$ is $h:W\to W'$ with $hq=q'$. This category is small because the skeleton is small and every relevant Hom is a small set. It is cofiltered. The map $V\to0$ supplies an object; two maps have a common predecessor through their product; and parallel $h,h'$ equal on $q(V)$ are equalized by factoring $q$ through the finite-dimensional kernel of $h-h'$.

Canonical double-dual interface. The limit of the diagram sending $q:V\to W$ to $W$ is naturally $V^{**}$. Its projection at $q$ is

p_q=\delta_W^{-1}q^{**}:V^{**}\longrightarrow W. \tag{1.3}

This limit is the underlying construction in Leinster, Theorem 7.5, which also identifies its codensity monad. We prove the required limit and its maps directly. Compatible choices $w_q\in W$ correspond to $t\in V^{**}$ by

\begin{gathered} t(\lambda)=w_{\lambda:V\to k},\\ \beta(w_q)=t(\beta q)\quad(\beta\in W^*). \end{gathered} \tag{1.4}

To prove linearity, take $\lambda,\nu\in V^*$ and the object $q=(\lambda,\nu):V\to k^2$. Compatibility with its two coordinate maps makes $w_q=(t(\lambda),t(\nu))$. Compatibility with the linear map $(a,b)\mapsto a+b$ then gives $t(\lambda+\nu)=t(\lambda)+t(\nu)$. Compatibility with multiplication by any $c\in k$ gives $t(c\lambda)=ct(\lambda)$, including $c=0$. The use of $k$ and $k^2$ is legitimate on the chosen skeleton by transporting them along their chosen isomorphisms. Thus $t$ is a linear functional on $V^*$. For any $q:V\to W$, compatibility with each $\beta:W\to k$ gives the second equality in (1.4). Finite-dimensional evaluation therefore forces $w_q=\delta_W^{-1}q^{**}(t)$.

Conversely every $t\in V^{**}$ gives the compatible family (1.3): if $hq=q'$, naturality of finite-dimensional evaluation gives $hp_q=p_{q'}$. Evaluating its component at each $\lambda$ returns $t(\lambda)$, so the two constructions are inverse. Both are linear, component by component. The compatible tuples form the actual vector-space limit, a linear subspace of the product, and this proves the asserted isomorphism and universal property. For $f:V\to U$, precomposition sends the component at $q:U\to W$ to that at $qf$; formula (1.4) identifies the resulting map with $f^{**}$. The cone from $V$ is precisely $\delta_V$, since $p_q\delta_V=q$.

The index includes every finite-dimensional target map, rather than a selected coordinate tower. A smaller family of quotients can have a different inverse limit; Exercise 2 measures that distinction.

2. Density in the original direction

The restricted nerve is

\begin{gathered} N:\mathsf V\longrightarrow\widehat{\mathsf F},\\ N(V)(W)=\operatorname{Hom}_k(W,V). \end{gathered} \tag{2.1}

The single probe $k^2$ is dense by the finite free-probe theorem in the prerequisite lesson, specialized to $R=k$ and rank two. The enlargement lemma therefore proves that all of $\mathsf F$ is dense. In particular, every set-valued natural transformation $N(V)\to N(U)$ is induced by exactly one linear map $V\to U$. Additivity is forced by probe operations; it need not be imposed on the components of the transformation.

Now put $\mathsf C=\mathsf V^{\rm op}$ and $\mathsf A=\mathsf F^{\rm op}$. Its nerve is a covariant functor on the original finite-dimensional category:

\begin{gathered} N_{\rm op}:\mathsf V^{\rm op}\longrightarrow \mathsf{Set}^{\mathsf F},\\ N_{\rm op}(V)(W)=\operatorname{Hom}_k(V,W). \end{gathered} \tag{2.2}

For finite-dimensional $W$, transpose gives

\operatorname{Hom}_k(V,W) \simeq\operatorname{Hom}_k(W^*,V^*). \tag{2.3}

The inverse sends $g:W^*\to V^*$ to the map taking $v$ to the unique $w\in W$ such that $\beta(w)=g(\beta)(v)$ for all $\beta\in W^*$. Existence and uniqueness use $\delta_W$; linearity follows by evaluating against every $\beta$. Precomposition in $V$ and postcomposition in $W$ verify both-variable naturality directly.

Let $E:\widehat{\mathsf F}\to\mathsf{Set}^{\mathsf F}$ send $P$ to $W\mapsto P(W^*)$. Finite duality makes $E$ an equivalence. Equation (2.3) is the natural factorization

N_{\rm op}\simeq E\,N\,(-)^*. \tag{2.4}

Thus reversing the probes exposes the fullness of algebraic dualization. The two equivalences just used concern finite-dimensional objects and presheaf categories; they do not say that dualization is an equivalence on all vector spaces.

3. An explicit natural transformation that is not a map

Let $V=k^{(\mathbb N)}$ be the direct sum with basis $e_1,e_2,\ldots$. Its algebraic dual is the full product $k^{\mathbb N}$: a functional is specified by an arbitrary sequence of values on the basis. Let $S\subset k^{\mathbb N}$ be the finite-support subspace, and let $\mathbf1=(1,1,\ldots)$.

Since $\mathbf1\notin S$, the rule

t(s+c\mathbf1)=c \quad(s\in S,\ c\in k) \tag{3.1}

defines a linear functional on $S\oplus k\mathbf1$. Extend it to $k^{\mathbb N}$ using a basis. Then $t\in V^{**}$, but $t\notin\delta_V(V)$. Indeed, if $t=\delta_V(v)$, evaluating on the coordinate sequences in $S$ makes every coefficient of $v$ zero. This contradicts $t(\mathbf1)=1$. The argument works over every field.

A morphism $k\to V$ in $\mathsf V$, viewed as a morphism $V\to k$ in $\mathsf V^{\rm op}$, is carried by duality to a map $V^*\to k$. Its value is evaluation at the image of $1$. The map (3.1) is therefore outside the image of this Hom map: dualization $\mathsf V^{\rm op}\to\mathsf V$ is not full.

There is also a direct probe description. For $W\in\mathsf F$ and $q:V\to W$, define

\begin{gathered} \sigma_W(q):k\longrightarrow W,\\ \sigma_W(q)(1)=\delta_W^{-1}q^{**}(t). \end{gathered} \tag{3.2}

Naturality in $W$ follows from the compatible projections (1.3), so $\sigma:N_{\rm op}(V)\to N_{\rm op}(k)$ is a natural transformation of set-valued presheaves on $\mathsf F^{\rm op}$. If it were induced by $k\to V$, its component at $W=k$ would be evaluation at one vector of $V$. That would make $t=\delta_V(v)$, already disproved. Hence $\mathsf F^{\rm op}$ is not dense in $\mathsf V^{\rm op}$.

4. The missing base-change hypothesis

The strict-quotient density criterion requires small colimits and finite limits, stability of filtered colimits under base change, every epimorphism strict, and a finite-coproduct-closed detecting family. For $(\mathsf V^{\rm op},\mathsf F^{\rm op})$, all conditions except base-change stability hold.

Vector spaces have arbitrary limits, obtained from compatible tuples in products, and arbitrary colimits, obtained as quotients of direct sums by the arrow relations. Thus their opposite has both required kinds of constructions. Finite coproducts in the opposite are finite products of vector spaces, and finite-dimensional spaces are closed under them.

The object $k$ detects isomorphisms in the opposite. For a linear map $f:X\to Y$, its dual is injective exactly when $f$ is surjective, by functionals on the cokernel. It is surjective exactly when $f$ is injective: an injective map allows extension of every functional, while a nonzero kernel vector supports a functional that cannot be in the image of $f^*$. Consequently bijectivity of $f^*$ implies that $f$ is invertible. These are the componentwise probe tests.

An epimorphism in $\mathsf V^{\rm op}$ corresponds to a monomorphism $i:X\to Y$ in $\mathsf V$. It is strict in the opposite because $i$ equalizes its self-pushout in vector spaces. Explicitly, the pushout is

(Y\oplus Y)/\{(i(x),-i(x)):x\in X\}.

The two maps from $Y$ agree at $y$ precisely when $y\in i(X)$. Thus their equalizer is exactly $i$, the dual strictness condition. This is the ordinary module interface from the prerequisite lesson.

Here is an actual filtered-base-change failure. Let $B=k[z]$, and for $n\ge1$ put $Y_n=B/(z^n)$. Quotient maps $Y_{n+1}\to Y_n$ give an inverse system in $\mathsf V$, hence a filtered diagram in $\mathsf C=\mathsf V^{\rm op}$. The quotient maps $B\to Y_n$ make it a diagram over $B$ in $\mathsf C$.

Compatible truncated polynomials are exactly arbitrary coefficient sequences. Therefore

\lim_nY_n=k[[z]] \tag{4.1}

as vector spaces, and the induced map $B\to\lim_nY_n$ includes polynomials as the eventually zero coefficient sequences. This uses formal series only.

Base change in $\mathsf C$ along $0\to B$ corresponds to pushout in $\mathsf V$ along $B\to0$. Each stage gives

Y_n\amalg_B0=\operatorname{coker}(B\to Y_n)=0. \tag{4.2}

At the colimit stage in $\mathsf C$, however, the same pullback is the opposite of

k[[z]]\amalg_B0=k[[z]]/k[z]\ne0. \tag{4.3}

The class of $\sum_{j\ge0}z^j$ is nonzero over every field. The canonical comparison from the colimit of the stage pullbacks to the pullback of the colimit is thus $0\to k[[z]]/k[z]$ in $\mathsf C$; in vector-space direction it is the map $k[[z]]/k[z]\to0$. It is not invertible. This identifies the failed hypothesis, with the arrows and the nonzero obstruction specified.

5. Graded exercises with complete solutions

Exercise 1 — Foundation: annihilators and exact duals. Let $U\subset V$, and let $U^\perp\subset V^*$ consist of functionals vanishing on $U$. Prove $0\longrightarrow(V/U)^* \longrightarrow V^*\longrightarrow U^* \longrightarrow0$ is exact. Prove that $U$ is the set of vectors annihilated by every element of $U^\perp$. No finite-dimensional assumption is allowed.

Solution. Pullback along $V\to V/U$ is injective and has image exactly $U^\perp$, by the quotient universal property. Restriction $V^*\to U^*$ has that kernel. It is surjective because a basis of $U$ extends to one of $V$, and any functional can be assigned arbitrary, for example zero, values on the additional basis elements. If $v\notin U$, its class in $V/U$ is nonzero; a coordinate functional on a basis containing that class does not vanish on it. Pulling it back gives an element of $U^\perp$ that does not annihilate $v$. Every element of $U$ is annihilated by definition. This proves both assertions, including $U=0$ and $U=V$.

Exercise 2 — Intermediate: a coordinate tower misses information. For $V=k^{(\mathbb N)}$, let $q_n:V\to k^n$ retain the first $n$ coordinates. Identify $\lim_n k^n$, and describe the canonical restriction $r:V^{**}\longrightarrow\lim_n k^n$ from the full finite-target limit in Section 1. Prove that $r$ is surjective and has a nonzero kernel.

Solution. Compatible tuples of initial segments are arbitrary sequences, so the displayed limit is $k^{\mathbb N}$. Identify $V^*=k^{\mathbb N}$, and write $\varepsilon_j\in V^*$ for its $j$-th coordinate sequence. Formula (1.3) gives $r(t)=(t(\varepsilon_1),t(\varepsilon_2),\ldots).$ For any sequence $c=(c_j)$, prescribing $t(\varepsilon_j)=c_j$ defines a functional on the span of the $\varepsilon_j$, since each linear combination is finite. Basis extension extends it to $V^*$, proving surjectivity. The functional (3.1) vanishes on every $\varepsilon_j$ and is nonzero, so it lies in $\ker r$. More precisely, the kernel consists of functionals vanishing on the finite-support subspace of $V^*$, hence identifies with $(V^*/S)^*$. The coordinate tower recovers a quotient of the full completion and cannot replace its entire probe index.

Exercise 3 — Advanced: completion detects this base-change obstruction. Let $K_1\supset K_2\supset\cdots$ be subspaces of a vector space $B$. Form the inverse quotient system $B/K_n$, its limit $L$, and the canonical map $c:B\to L$. In $\mathsf V^{\rm op}$, show that base change along $0\to B$ commutes with this particular filtered colimit exactly when $c$ is surjective. Also identify $\ker c$, and decide the condition for a finite-dimensional $B$.

Solution. Every stage is a quotient of $B$, so its ambient pushout along $B\to0$ is zero. The ambient pushout at the limit is $L/c(B)$. Thus the opposite comparison is invertible exactly when this quotient is zero, equivalently when $c$ is surjective. Its kernel is $\bigcap_nK_n$, since a vector has zero class at every stage exactly under that condition. If $B$ is finite-dimensional, the nonincreasing integer dimensions of the $K_n$ stabilize. Nested subspaces of equal finite dimension coincide, so from some stage onward all quotients and transitions are constant isomorphisms. Then $L=B/K_N$ for such a stage, and $c$ is surjective. Section 4 exhibits an infinite-dimensional $B$ where surjectivity fails. An injective $c$ alone would not settle the base-change comparison.

Exercise 4 — Expert: characterize exactly which transpose maps come from maps. Given any vector spaces $V,W$ and a linear map $a:W^*\to V^*$, prove that it is $f^*$ for a unique linear $f:V\to W$ exactly when $a^*(\delta_V(V))\subset\delta_W(W). \tag{5.1}$ Then prove that $\delta_V$ is surjective exactly when $V$ is finite-dimensional.

Solution. If $a=f^*$, naturality of evaluation says $a^*\delta_V=\delta_Wf$, proving the inclusion. Conversely, since $\delta_W$ is injective, (5.1) defines $f(v)=\delta_W^{-1}\bigl(a^*\delta_V(v)\bigr).$ This is linear, as all maps in the formula are linear on the indicated subspace. For $\lambda\in W^*$, $\lambda(f(v))=a(\lambda)(v),$ so $f^*=a$. Functionals separate vectors of $W$, proving uniqueness.

Finite-dimensional surjectivity was proved in Section 1. If $V$ is infinite-dimensional, choose countably many elements of a basis. Their span $V_0$ is isomorphic to $k^{(\mathbb N)}$, and the remaining basis elements give a complement. Write $i:V_0\to V$ and $p:V\to V_0$ for inclusion and projection, so $pi=1$. If $\delta_V$ were surjective, take any $t\in V_0^{**}$ and choose $v\in V$ with $\delta_V(v)=i^{**}(t)$. Naturality then gives $\delta_{V_0}(p(v)) =p^{**}\delta_V(v) =p^{**}i^{**}(t)=t.$ That would make $\delta_{V_0}$ surjective, contradicting (3.1). Thus every infinite-dimensional $V$ fails surjectivity. The argument uses no cardinal comparison of a space and its dual and works uniformly for finite and infinite fields.

Self-checked; no independent review has occurred. Original exposition and exercise text are dedicated to the public domain under CC0 1.0.