Does the tensor product preserve coreflexive equalizers? 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 $f,g:M\to N$ such that there exists $r:N\to M$ satisfying $r\circ f=\mathsf{id}_M=r\circ g$. By a coreflexive equalizer I mean the equalizer of a coreflexive pair.

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

I know that $P\otimes -$ does not preserve equalizers in general, but a coreflexive equalizer is something more particular.
 A: There is always a natural map $\eta : \ker(f-g)\otimes P\to \ker((f-g)\otimes P)$, 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 $\mathbb Z/2\otimes a = 0$. 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 $f:M\to N$ and let $F : \widetilde{M}\to N\oplus\widetilde M$ 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 $\ker f\oplus \widetilde M$.
Then you obtain that the map $K\otimes P\to \ker((F-G)\otimes P)$ is just the map with coordinates the natural map $\ker f\otimes P\to \ker(f\otimes P)$ and the identity of $\widetilde M\otimes P$, so the claim follows. 
