Proposition. Finite-map pullback and fiber weights [fdip-0028]
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  • October 23, 2026

Let \(\rho :\Omega \to \Lambda \) be any map of finite sets and let \(R\) be a commutative unital ring. On the function algebras, \[\begin {aligned} \rho ^*:A_\Lambda &\longrightarrow A_\Omega ,& f&\longmapsto f\circ \rho \end {aligned}\] is a unital \(R\)-algebra map. Over \(\mathbb R\) or \(\mathbb C\) it preserves pointwise conjugation. For \(\tau :\Lambda \to \Gamma \), \((\tau \circ \rho )^*=\rho ^*\circ \tau ^*\), and the identity set map induces the identity algebra map.

Proof. 

  • October 23, 2026

At \(x\in \Omega \), pullback of \(rf+sg\), \(fg\), and \(1\) has value \(rf(\rho (x))+sg(\rho (x))\), \(f(\rho (x))g(\rho (x))\), and \(1_R\), respectively. Pullback commutes with conjugation by the same evaluation. Both composite pullbacks evaluate \(f\) at \(\tau (\rho (x))\); the identity case evaluates \(f\) at \(x\). These pointwise arguments also apply to empty domains and the zero ring.

The image of \(\rho ^*\) consists exactly of functions constant on each fiber of \(\rho \). If \(\rho \) is surjective, \(\rho ^*\) is injective. Conversely, injectivity of \(\rho ^*\) forces surjectivity of \(\rho \) provided \(R\) is nonzero.

Proof. 

  • October 23, 2026

A pullback takes equal values at \(x,x'\) whenever \(\rho (x)=\rho (x')\). Conversely, for fiber-constant \(g\), define \(f(y)\) to be the common value on the fiber when that fiber is nonempty, and \(0_R\) when it is empty. The value is independent of the chosen representative, and \(g=f\rho \). If \(\rho \) is surjective, equality of two pullbacks determines their value at every \(y\), proving injectivity. If \(y\) is missed, \(\delta _y\) pulls back to zero; for nonzero \(R\) this indicator is nonzero, so pullback is not injective. Over the zero ring, all these algebras are singletons and every pullback is injective, including pullback along a nonsurjective set map.

For the weight functional determined by \(p\), its pullback as a functional along \(\rho ^*\) has weights \[q(y)=\sum _{\rho (x)=y}p(x).\] Thus normalized weights stay normalized; real nonnegative weights stay nonnegative. Empty fibers have weight zero.

Proof. 

  • October 23, 2026

Regroup the finite sum: \[\begin {aligned} (\omega _p\circ \rho ^*)(f) &=\sum _{x\in \Omega }p(x)f(\rho (x))\\ &=\sum _{y\in \Lambda }\left (\sum _{\rho (x)=y}p(x)\right )f(y). \end {aligned}\] The fibers partition \(\Omega \), so \(\sum _yq(y)=\sum _xp(x)\). Nonnegative summands give nonnegative fiber sums. For successive maps, \[\sum _{\tau (y)=z}\sum _{\rho (x)=y}p(x) =\sum _{\tau (\rho (x))=z}p(x),\] which proves compatibility of the fiber formula with composition.

For real nonnegative weights of total one, this is pushforward of finite probabilities, or a marginal law when \(\rho \) forgets coordinates. It uses no division by fiber mass, even when that mass is zero; it is not a conditional expectation. The pointwise algebra construction uses [stacks2017stacks, Section 22.2], Tag 09JF. Mathlib's AlgHom.pi constructs the family of evaluation maps \(f\mapsto f(\rho (x))\). Composition here changes the function's domain, not its coefficient ring.

Over arbitrary rings, not every unital algebra map between function algebras is induced by a set map. For \(R=\mathbb Z/6\mathbb Z\), define \(h:R^2\to R\) by \(h(a,b)=3a+4b\). It is \(R\)-linear and unital since \(3+4=1\). The identities \(3^2=3\), \(4^2=4\), and \(3\cdot 4=0\) show \(h(a,b)h(c,d)=3ac+4bd=h(ac,bd)\), so it is multiplicative. But \(h(1,0)=3\), whereas either coordinate evaluation gives \(0\) or \(1\). Thus this map is not the pullback of a map from a singleton to two points.