Vtyjkyuk

Let $Bbbk$ be any commutative ring and $otimes$ denote the tensor product over $Bbbk$. By a coreflexive pair I mean a parallel pair of $Bbbk$-module homomorphisms admitting a common retraction, i.e. a pair of arrows $f,g:Mto N$ such that there exists $r:Nto M$ satisfying $rcirc f=mathsfid_M=rcirc g$. By a coreflexive equalizer I mean the equalizer of a coreflexive pair.

Q: Does $Potimes -$ preserve coreflexive equalizers for every $Bbbk$-module $P$?

I know that $Potimes -$ does not preserve equalizers in general, but a coreflexive equalizer is something more particular.

There is always a natural map $eta : ker(f-g)otimes Pto ker((f-g)otimes P)$, but it is not always an isomorphism, as the following example shows.

Consider $M=N= bigoplus mathbb Z$ indexed over the naturals, and the maps $f(x) = (0,x_1,x_2,ldots)$, $g(x) = (2x_1,x_1,x_2,ldots)$, and $r(x) = (x_2,x_3,ldots)$, so that $rf=rg=1$ and $a = g-f$ is given by $a(x) = (2x_1,0,0,ldots)$. Then $K=ker a = langle (0,x_2,x_3,ldots)rangle$, while $mathbb Z/2otimes a = 0$. It follows that the map $eta$ is not an isomorphism.

Using this idea, you can prove that tensor products preserve equalizers iff they preserve the reflexive ones. Indeed, consider any map of $mathbb k$-modules $f:Mto N$ and let $F : widetildeMto Nopluswidetilde M$ be the map $(m_1,m_2,ldots)mapsto (fm_1,m_1,m_2,ldots)$, where the tilde means taking a sum. Let $G$ be the map where $f=0$. Then $F$ and $G$ admit a common retraction given by sending $(n,m_1,m_2,ldots)$ to $(m_1,m_2,ldots)$, and $K=ker(F-G)$ is $ker foplus widetilde M$.

Then you obtain that the map $Kotimes Pto ker((F-G)otimes P)$ is just the map with coordinates the natural map $ker fotimes Pto ker(fotimes P)$ and the identity of $widetilde Motimes P$, so the claim follows.