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In a question that I recently posted I defined a property of some concrete categories $(*)$, that if it holds, then:

A subobject of $X$ is uniquely determined by the underlying set (subset of $U(X)$), same for quotient object (namely in this case, determined by the equivalence relation on $U(X)$).

So now I will be referring only to concrete categories that satisfy this property, and I'm about to define yet another property. But first, motivation.

In the category $\textbf{Rng}$ (rings without identity) a morphism is a function that preserves addition and multiplication. But in the category $\textbf{Ring}$, a morphism is a rng-morphism, that also takes identity to identity. So I was wondering, what is the advantage in adding this condition?

Let's look at a ring as a set that is a group with respect to addition and a monoid with respect to multiplication. So we can make the question simpler: why do we define a monoid morphism as a function that preserves the operation and takes identity to identity?

So I claim that from this definition of morphism, we obtain that it's a category which satisfies "the first isomorphism theorem":

  1. For any morphism $f:X\to Y$, the image of $U(f)$ is a subobject of $U(Y)$ (I can say that a subset of the underlying set "is a subobject" exactly because I assume property $(*)$).
  2. Moreover, we can obviously look at the equivalence relation a function induces on its domain: $U(f)$ defines the relation $a_1\sim_{U(f)}a_2$ iff $Uf(a_1)=Uf(a_2)$. So the claim is: $U(X)/\sim_{U(f)}$ is a quotient object of $X$.
  3. The function between the underlying sets of those objects $Im(U(f))\to U(X)/\sim_{U(f)}, b\mapsto f^{-1}(b)$ is an isomorphism.

It's easy to check that the first isomorphism theorem holds in the category $\textbf{Mon}$ if and only if we define morphisms to take identity to identity. And it holds in $\textbf{Ring}$ if and only if we define morphisms to be both morphisms of the additive group and of the multiplicative monoid.

My take on it is, it's not like algebraists define these structures and then notice "hey, that satisfies the first isomorphism theorem" but rather define the structures specifically so that it would happen (obviously, this is nothing new). So idk if this even counts as a question, I'd like to know if people find mistakes what I wrote (I basically made that up), and also if property $(*)$ has an actual name.

rutruttt
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  • In what sense first isomorphism theorem holds for topological spaces? Most certainly a continous image of a space does not have to be its quotient. For example every space is an image of some discrete space, but a quotient of discrete is also discrete. – freakish Mar 16 '24 at 22:51
  • lol youre right i overexaggerated, thanks for the correction :) – rutruttt Mar 16 '24 at 23:06
  • From the point of view of universal algebra, a semigroup has one binary operation, a monoid has one binary operation and one nullary operation (corresponding to the identity), and morphisms are required to respect both operations. Groups have three operations: a binary (multiplication), a unary (inverses), and a nullary. It is then a theorem, rather than a definition, that semigroup morphisms between groups are actually group morphisms. Rngs have four operations: $+$ and $\cdot$ (binary), $-$ (unary), and $0$ (nullary). Rings have an additional nullary operation. $1$. – Arturo Magidin Mar 17 '24 at 01:17
  • The isomorphism theorems hold for any variety of $\Omega$ algebras. Here, $\Omega$ is the type (listing the arities of the operations), and "variety" means thatvthey satisfy a set of identities. – Arturo Magidin Mar 17 '24 at 01:19

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The first isomorphism theorem can be generalised as an axiom. It's one way to define Abelian category, actually. $\mathscr{A}$ is Abelian iff. it is additive, has all kernels and cokernels, and all morphisms are strict (first isomorphism theorem).

$\phi:a\to b$ is said to be strict if the induced map $\mathrm{coker}\ker\phi\to\ker\mathrm{coker}\,\phi$ is an isomorphism. That is, if $\mathrm{Coim}(\phi)\cong\mathrm{Im}(\phi)$, if $a/\mathrm{Ker}\,\phi\cong\mathrm{Im}(\phi)$, via the natural map induced by $\phi$. We don't care if they're abstractly isomorphic, we need them isomorphic in the canonical way.

FShrike
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