I've been reading the Queen Mary Note of Maurice Auslander 's paper Representation dimension of artin algebras. However I met some difficulties that are not so relevant to the topic. The following is such one.
On page six he describes the image of a morphism between two commutative rings in the category of rings with identities.
Let R and T be commutative rings and $f:R\rightarrow T$ a ring homomorphism. Let $f(R)=S$. We have $f_1:T\rightarrow T\otimes_ST$ defined by $f_1(t)=1\otimes t$ and $f_2(t)=t\otimes 1$.
Let $T'=\{t\in T| f_1(t)=f_2(t)\}$. Clearly $T'$ contains S, so that $f$ maps $R$ into $T'$. Prove that the inclusion map $S\rightarrow T'$ is an epimorphism in the category of rings and the inclusion map $T'\rightarrow T$ is a strong monomorphism.
I tried to use the notion dominion and pushout to prove the epimorphism part but failed.
The asserted claim is incorrect: the inclusion map $S\to T'$ need not be an epimorphism.
Note that there are categories of algebras (even varieties) where some subalgebras are properly contained in their dominion, but where epimorphisms are nonetheless surjective, so this claim would not provable from with a categorical approach: it isn't true even in varities of algebras.
But it is also not true in the case of commutative rings with $1$. I recently constructed a counterexample, not realizing the question has been asked here before.
Explicitly, let $T=\mathbb{Z}[x,y,z]$, and let $S=\mathbb{Z}[xy,y,yz]$. Then $T'=\mathbb{Z}+yT$. Indeed, if $x^ay^bz^c$ is any monomial divisible by $y$ ($b\gt 0$) which lies in $T'$, then so does $x^{a+1}y^bz^c$: if $f,g\colon T\to \mathfrak{T}$ are homomorphisms with $f|_S=f|_S$, then $$\begin{align*} f(x^{a+1}y^bz^c) &= f(x(x^ay^bz^c))\\ &=f(x)f(x^ay^bz^c)\\ &= f(x)g(x^ay^bz^c)\\ &= f(x)g(y)g(x^ay^{b-1}z^c)\\ &= f(x)f(y)g(x^ay^{b-1}z^c)\\ &= f(xy)g(x^ay^{b-1}z^c)\\ &= g(xy)g(x^ay^{b-1}z^c)\\ &= g(x^{a+1}y^bz^c). \end{align*}$$ A symmetric argument shows that if $x^ay^bz^c\in T'$ with $b\gt 0$, then $x^ay^bz^{c+1}\in T'$. Thus, every monomial divisible by $y$ lies in $T'$. Conversely, if $p(x,y,z)\notin \mathbb{Z}+yT$. then the maps $T\to T$ that maps $x\mapsto x$, $z\mapsto z$ and $y\mapsto 0$, and the one that maps $x,y,z\mapsto 0$ agree on $S$, but disagree on $p(x,y,z)$.
To show that the inclusion $S\to T'$ is not an epimorphism it suffices to show that there exist a commutative ring with $1$, $\mathfrak{T}$, and two morphism $T'\to \mathfrak{T}$ that agree on $S$ but not on all of $T'$. For example, $$ \mathfrak{T} = \mathbb{Z}[\epsilon]/(\epsilon^2),$$ the ring of dual numbers. Let $f,g\colon T'\to \mathfrak{T}$ be the morphisms defined by sending $xy$, $y$, and $yz$ to $\epsilon$; $f$ sends $xyz$ to $\epsilon$ and $g$ sends it to $0$. And both $f$ and $g$ send every other monomial to $0$. It is not hard to verify these maps are multiplicative, but since they disagree on $xyz$, the inclusion $S\to T'$ is not an epimorphism.