A functor that preserves binary coproducts or binary products between additive categories must preserve the zero object

additive-categories

That is the question. If I have a functor $F:\mathcal{A}\to\mathcal{B}$ between additive categories such that either

  1. $F(A)\oplus F(B)\to F(A\oplus B)$ is an isomorphism for all $A,B\in\mathcal{A}$, or
  2. $F(A\oplus B)\to F(A)\oplus F(B)$ is an isomorphism for all $A,B\in\mathcal{A}$

holds (i.e., such that it either preserves binary coproducts or binary products), then it must preserve the zero object, $F(0)= 0$?

Best Answer

The result is true.

Assume $F$ commutes with products (i.e., condition 1 holds). It suffices to verify that $F(0)$ is terminal (on an preadditive category, for an object, the conditions of being initial, terminal and the zero object are the equivalent). If $B\in\mathcal{B}$, then there is at least one map $B\to F(0)$ (the zero map), since $\mathcal{B}$ is preadditive. Now let $f,g:B\to F(0)$ be maps. The morphism $f-g$ equals the composite $$ B\xrightarrow{f\times g}F(0)\oplus F(0)\xrightarrow{p'_1-p'_2} F(0), $$ where $p_1',p_2':F(0)\oplus F(0)\to F(0)$ are the projections. Since $0\oplus 0=0$ is an initial object, the projections $p_1,p_2:0\oplus 0\to 0$ satisfy $p_1=p_2$. Thus, since the diagram

enter image description here

commutes, for $i=1,2$, and $Fp_1\times Fp_2$ is an isomorphism, it follows that $p_1'=p_2'$. Therefore $f-g=(p_1'-p_2')\circ(f\times g)=0\circ(f\times g)=0$. Hence, $F(0)$ is terminal in $\mathcal{B}$. Dually, with a similar argument, from condition 2 one can show that $F(0)$ is an initial object.

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