Infinite products and coproducts in additive categories

In a category with zero morphisms, there is a canonical morphism ⨁i∈IXi→∏i∈IXi\bigoplus_{i \in I} X_i \to \prod_{i \in I} X_i whenever the coproduct and product exist. In most categories, if II is infinite and the objects XiX_i are non-zero, this morphism is not an isomorphism. The following result generalizes this observation.

Proposition.

Let C\C be an additive category with countable powers and countable copowers. Let A∈CA \in \C be an object such that the canonical morphism α:⨁n≥1A→∏n≥1A\textstyle \alpha : \bigoplus_{n \geq 1} A \to \prod_{n \geq 1} A is an isomorphism. Then A=0A = 0.

Proof. Define the diagonal morphism Δ:A→∏n≥1A\textstyle \Delta : A \to \prod_{n \geq 1} A by pn∘Δ=idAp_n \circ \Delta = \id_A, and the codiagonal morphism ∇:⨁n≥1A→A\textstyle \nabla : \bigoplus_{n \geq 1} A \to A by ∇∘in=idA\nabla \circ i_n = \id_A for all n≥1n \geq 1. Define the endomorphism x∈End⁡(A)x \in \End(A) by x≔∇∘α−1∘Δ.x \coloneqq \nabla \circ \alpha^{-1} \circ \Delta.

Using the decompositions ⨁n≥1A≅A⊕⨁n≥2A,\textstyle \bigoplus_{n \geq 1} A \cong A \oplus \bigoplus_{n \geq 2} A, ∏n≥1A≅A⊕∏n≥2A,\textstyle \prod_{n \geq 1} A \cong A \oplus \prod_{n \geq 2} A, the morphisms α\alpha, Δ\Delta, and ∇\nabla have the following block matrix forms:

  • α=(idA00α′)\alpha = \begin{pmatrix} \id_A & 0 \\ 0 & \alpha' \end{pmatrix}, where α′:⨁n≥2A→∏n≥2A\alpha' : \bigoplus_{n \geq 2} A \to \prod_{n \geq 2} A is the canonical morphism.
  • Δ=(idAΔ′)\Delta = \begin{pmatrix} \id_A \\ \Delta' \end{pmatrix}, where Δ′:A→∏n≥2A\Delta' : A \to \prod_{n \geq 2} A is the diagonal morphism.
  • ∇=(idA∇′)\nabla = \begin{pmatrix} \id_A & \nabla' \end{pmatrix}, where ∇′:⨁n≥2A→A\nabla' : \bigoplus_{n \geq 2} A \to A is the codiagonal morphism.

Since α\alpha is an isomorphism, α′\alpha' is an isomorphism as well. We compute:

x=∇∘α−1∘Δ=(idA∇′)∘(idA00α′−1)∘(idAΔ′)=(idA∇′)∘(idAα′−1∘Δ′)=idA+∇′∘α′−1∘Δ′. \begin{align*} x & = \nabla \circ \alpha^{-1} \circ \Delta \\ & = \begin{pmatrix} \id_A & \nabla' \end{pmatrix} \circ \begin{pmatrix} \id_A & 0 \\ 0 & \alpha'^{-1} \end{pmatrix} \circ \begin{pmatrix} \id_A \\ \Delta' \end{pmatrix} \\ & = \begin{pmatrix} \id_A & \nabla' \end{pmatrix} \circ \begin{pmatrix} \id_A \\ \alpha'^{-1} \circ \Delta' \end{pmatrix} \\ & = \id_A + \nabla' \circ \alpha'^{-1} \circ \Delta'. \end{align*}

Now define the isomorphism σ:⨁n≥1A→⨁n≥2A\textstyle \sigma : \bigoplus_{n \geq 1} A \to \bigoplus_{n \geq 2} A by σ∘in=in+1\sigma \circ i_n = i_{n+1} for n≥1n \geq 1. Likewise, define the isomorphism τ:∏n≥1A→∏n≥2A\textstyle \tau : \prod_{n \geq 1} A \to \prod_{n \geq 2} A by pn∘τ=pn−1p_n \circ \tau = p_{n-1} for n≥2n \geq 2.

By the definitions of α′\alpha', Δ′\Delta', and ∇′\nabla', we have

α′=τ∘α∘σ−1,Δ′=τ∘Δ,∇′=∇∘σ−1. \begin{align*} \alpha' & = \tau \circ \alpha \circ \sigma^{-1}, \\ \Delta' & = \tau \circ \Delta, \\ \nabla' & = \nabla \circ \sigma^{-1}. \end{align*}

Therefore,

∇′∘α′−1∘Δ′=∇∘σ−1∘(τ∘α∘σ−1)−1∘τ∘Δ=∇∘σ−1∘σ∘α−1∘τ−1∘τ∘Δ=∇∘α−1∘Δ=x. \begin{align*} \nabla' \circ \alpha'^{-1} \circ \Delta' & = \nabla \circ \sigma^{-1} \circ (\tau \circ \alpha \circ \sigma^{-1})^{-1} \circ \tau \circ \Delta \\ & = \nabla \circ \sigma^{-1} \circ \sigma \circ \alpha^{-1} \circ \tau^{-1} \circ \tau \circ \Delta \\ & = \nabla \circ \alpha^{-1} \circ \Delta \\ & = x. \end{align*}

Thus x=idA+xx = \id_A + x, and hence idA=0\id_A = 0. Therefore A=0A = 0. □\square

Context

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