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I'm currently on my second semester of Abstract Algebra, and one thing that has bothered me recently is having to prove the range of an application (that we wish to prove is an epimorphism) is an algebra / ring / (any other common structure).

Suppose we have an algebraic structure of some type $R$, with $n$ operations that satisfy some axioms like associativity or commutativity, and possible identities. Does it not follow that if $S$ is a set with $n$ corresponding operations (not assuming that they satisfy the axioms or have corresponding identities), and that there exists some surjective application which preserves the operations $f:R \to S$ then $S$ is the same type of algebraic structure as $R$, the identities of $S$ are the images of the identities of $R$ and $f$ is an epimorphism?

For a specific algebraic structure: If $R$ is a ring, and $S$ is a set with two binary applications $\times: S \times S \to S$ and $+: S\times S \to S$ such that there exists a surjective application $f: R \to S$ with $f(x+y) = f(x)+f(y)$ and $f(xy) = f(x)f(y)$ then is it true that $S$ is a ring and $f$ is a ring epimorphism?

TC159
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The type of an algebraic structure is the collection of arities of its operations. Since you have $n$ operations in $R$ and $S$ and they must correspond via preservation of the operations under $f$, the types are necessarily the same (up to re-indexing of the operations in $S$).

To see this, just write out such a preservation property: $$\forall r_1,...,r_{n_i}\in R:\text{ }\text{ }f\bigg(o_i\big(r_1,...,r_{n_i}\big)\bigg) = \tilde{o}_j\big(f(r_1),...,f(r_{n_i})\big),$$ where $o_i$ and $\tilde{o}_j$ are $n_i$-ary operations in $R$ and $S$ respectively. We just reindex $j\leftrightarrow i$.

If we consider identities as unary operations and append the type, we have: $$\forall r\in R:\text{ }f(1_R(r))=1_S(f(r))$$ so that identities map to identities. It should be noted that one usually says homomorphism of structures of type $\tau$ (or -morphism of groups, rings, etc.), which indicates the types are the same.

To clarify, such a map need not preserve relations defined on $R$ (this is a separate thing). In abstract algebra especially, the relations are tied to our notion of type (Group, Ring, etc.). But in the same way, if there are $m$ amount of relations on both and we have a correspondence as above, the "type" by extension should be the same.

An epimorphism is a term from category theory (not unrelated to universal algebra, but a little different in flavor). These don't always correspond to surjective morphisms. Assuming $Alg(\tau)$ defines a category with appropriate morphisms just described, an epimorphism then is a morphism $f\in Hom(R,S)$ such that: $$\forall \varphi,\psi\in Hom(S,T):\text{ }[\varphi\circ f = \psi \circ f]\implies [\varphi = \psi].$$ So let us now take $f$ to be surjective and take arbitrary $\varphi,\psi \in Hom(S,T)$ with: $$\varphi\circ f = \psi\circ f.$$ We have $\forall s\in S,\exists r\in R: f(r)=s$.

Being more pedantic, $\forall s_0\in S$, $\exists r_0\in R: f(r_0)=s_0.$ This implies: $$\varphi(s_0) = \varphi(f(r_0)) = \psi(f(r_0)) = \psi(s_0).$$ Hence $\varphi = \psi$ identically and we're done. So in $Alg(\tau)$, surjective implies epic. $\square$

Kevin
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  • The part on relations was a bit unsatisfactory to me, but in generality is hard to say. I think in absence of relations on the codomain, we can induce them somehow from the domain through the homomorphic properties. For example in abelian groups, let us define an inverse relation: $(g,h)\in R\text{ }\leftrightarrow\text{ }gh = 1$. Then for $f\in Hom(G,H)$, we have: $1_H = f(1_G) = f(gh) = f(g)f(h)$ hence we may define $\big(f(g),f(h)\big)\in \tilde{R}$ on the codomain. – Kevin Mar 16 '23 at 04:36