3 fix a couple of tensors; a bit less dubious in the final paragraph

I think what you want is exactly that the dominion (in the sense of Isbell) of $B$ in $A$ be equal to $B$.

Recall that if you are in a category of algebras (in the sense of Universal Algebra), and $B\subseteq A$ is a subalgebra of $A$, then the dominion of $B$ in $A$ (relative to the category of context) is the set $$\{a\in A\mid \forall C\forall f,g\colon A\to C (f|_B=g|_B\implies f(a)=g(a))\}.$$

If $x$ lies in the dominion of $B$ in $A$, then the fact that the two embeddings of $A$ into $A\otimes A\otimes_B A$ agree on $B$ implies that they agree on $x$; that is, $x\otimes 1=1\otimes x$. Conversely, suppose that the two embeddings of $A$ into $A\otimes A\otimes_B A$ agree on $x$; if $C$ is any commutative ring and $f,g\colon A\to C$ are two maps that agree on $B$, then the universal property of $A\otimes_B A$ guarantees a homomorphism $\Phi\colon A\otimes A\otimes_B A\to C$ such that $f = \Phi\circ\lambda$ and $g=\Phi\circ\rho$, where $\lambda$ and $\rho$ are the left and right embeddings; then since $\lambda(x)=\rho(x)$, we conclude that $f(x)=g(x)$. Hence any element of the equalizer is in th dominion.

A characterization of dominions in the category of commutative rings is given in the Isbell-Mazet-Silver Zigzag Theorem: the dominion of $B$ in $A$ consists precisely of the elements of $A$ that can be written in the form $XYZ$, where $X$ is a row matrix, $Z$ is a column matrix, $Y$ is a square matrix of the appropriate size, all $X$, $Y$, and $Z$ have entries in $A$, and $XY$ and $YZ$ have entries in $B$.

So the inclusion of $B$ is the equalizer of the left and right embeddings if and only if $B$ is equal to its own dominion in $A$.
(Certainly, if Essentially, when the category is right-closed, the dominion of $B$ is not dominion-closed in $A$, then A$is always equal to the equalizer of the two embeddings$A\to A\amalg_{B}A$; in the category of commutative rings the tensor product functions as a binary coproduct, hence the dominion of$B$in$A$is not the equalizer of the two embeddings$A\to A\otimes_B A$; I guess I'm but this description does not actually answer your question, it just asks it in a little wary of different language; so the converse...actual answer you want is contained in the Isbell-Mazet-Silver Zigzag Theorem, when all is said and done.) 2 Mazet, not Mazur I think what you want is exactly that the dominion (in the sense of Isbell) of$B$in$A$be equal to$B$. Recall that if you are in a category of algebras (in the sense of Universal Algebra), and$B\subseteq A$is a subalgebra of$A$, then the dominion of$B$in$A$(relative to the category of context) is the set $$\{a\in A\mid \forall C\forall f,g\colon A\to C (f|_B=g|_B\implies f(a)=g(a))\}.$$ If$x$lies in the dominion of$B$in$A$, then the fact that the two embeddings of$A$into$A\otimes A$agree on$B$implies that they agree on$x$; that is,$x\otimes 1=1\otimes x$. Conversely, suppose that the two embeddings of$A$into$A\otimes A$agree on$x$; if$C$is any commutative ring and$f,g\colon A\to C$are two maps that agree on$B$, then the universal property of$A\otimes_B A$guarantees a homomorphism$\Phi\colon A\otimes A\to C$such that$f = \Phi\circ\lambda$and$g=\Phi\circ\rho$, where$\lambda$and$\rho$are the left and right embeddings; then since$\lambda(x)=\rho(x)$, we conclude that$f(x)=g(x)$. A characterization of dominions in the category of commutative rings is given in the Isbell-Mazur-Silver Isbell-Mazet-Silver Zigzag Theorem: the dominion of$B$in$A$consists precisely of the elements of$A$that can be written in the form$XYZ$, where$X$is a row matrix,$Z$is a column matrix,$Y$is a matrix of the appropriate size, all have entries in$A$,$XY$and$YZ$have entries in$B$. (See also this previous post.) So the inclusion of$B$is the equalizer of the left and right embeddings if and only if$B$is equal to its own dominion in$A$. (Certainly, if$B$is not dominion-closed in$A$, then$B$is not the equalizer; I guess I'm a little wary of the converse...) 1 I think what you want is exactly that the dominion (in the sense of Isbell) of$B$in$A$be equal to$B$. Recall that if you are in a category of algebras (in the sense of Universal Algebra), and$B\subseteq A$is a subalgebra of$A$, then the dominion of$B$in$A$(relative to the category of context) is the set $$\{a\in A\mid \forall C\forall f,g\colon A\to C (f|_B=g|_B\implies f(a)=g(a))\}.$$ If$x$lies in the dominion of$B$in$A$, then the fact that the two embeddings of$A$into$A\otimes A$agree on$B$implies that they agree on$x$; that is,$x\otimes 1=1\otimes x$. Conversely, suppose that the two embeddings of$A$into$A\otimes A$agree on$x$; if$C$is any commutative ring and$f,g\colon A\to C$are two maps that agree on$B$, then the universal property of$A\otimes_B A$guarantees a homomorphism$\Phi\colon A\otimes A\to C$such that$f = \Phi\circ\lambda$and$g=\Phi\circ\rho$, where$\lambda$and$\rho$are the left and right embeddings; then since$\lambda(x)=\rho(x)$, we conclude that$f(x)=g(x)$. A characterization of dominions in the category of commutative rings is given in the Isbell-Mazur-Silver Zigzag Theorem: the dominion of$B$in$A$consists precisely of the elements of$A$that can be written in the form$XYZ$, where$X$is a row matrix,$Z$is a column matrix,$Y$is a matrix of the appropriate size, all have entries in$A$,$XY$and$YZ$have entries in$B$. (See also this previous post.) So the inclusion of$B$is the equalizer of the left and right embeddings if and only if$B$is equal to its own dominion in$A$. (Certainly, if$B$is not dominion-closed in$A$, then$B\$ is not the equalizer; I guess I'm a little wary of the converse...)