Finite structures usually have no sequential colimits

Lemma.

Let C\C be a category with finite powers, including a terminal object 11. Let a:1→Xa : 1 \to X be a morphism. Assume that the sequence of morphisms (Xn,a):Xn→Xn+1(X^n,a) : X^n \to X^{n+1} for n≥0n \geq 0 admits a colimit (in:Xn→C)(i_n : X^n \to C). Then for every m≥0m \geq 0 there is a split epimorphism C→XmC \to X^m. In particular, if U:C→SetU : \C \to \Set is a functor preserving finite powers and card⁡(U(X))≥2\card(U(X)) \geq 2, then U(C)U(C) is infinite.

Proof. Let m≥0m \geq 0 be fixed. For n≥0n \geq 0 we define a morphism un:Xn→Xmu_n : X^n \to X^m as follows: It is the projection on the first mm factors for m≤nm \leq n, and (Xn,am−n)(X^n,a^{m-n}) for m≥nm \geq n (for m=nm=n these agree). With generalized elements this says: un(x1,…,xn)={(x1,…,xm)m≤n(x1,…,xn,a,…,a)m≥nu_n(x_1,\dotsc,x_n) = \begin{cases} (x_1,\dotsc,x_m) & m \leq n \\ (x_1,\dotsc,x_n,a,\dotsc,a) & m \geq n \end{cases} We claim that un=un+1∘(Xn,a)u_n = u_{n+1} \circ (X^n,a), i.e. un(x1,…,xn)=un+1(x1,…,xn,a).u_n(x_1,\dotsc,x_n) = u_{n+1}(x_1,\dotsc,x_n,a). If m≤nm \leq n (hence, m≤n+1m \leq n+1), both sides are equal to (x1,…,xm)(x_1,\dotsc,x_m). If m>nm > n, i.e. m≥n+1m \geq n+1, both sides are equal to (x1,…,xn,a,…,a)(x_1,\dotsc,x_n,a,\dotsc,a). This proves the claim.

Hence, there is a unique morphism φ:C→Xm\varphi : C \to X^m such that φ∘in=un\varphi \circ i_n = u_n for all n≥0n \geq 0. Since umu_m is the identity, φ\varphi is a split epimorphism. If UU is a functor with the mentioned properties, U(φ)U(\varphi) is also a split epimorphism from U(C)U(C) to U(Xm)≅U(X)mU(X^m) \cong U(X)^m, and U(X)mU(X)^m has ≥2m\geq 2^m elements. This holds for all mm, so that U(C)U(C) is infinite. □\square

Author: Martin Brandenburg

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