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When we define a group homomorphism $\theta \colon G \to H$, we do not have to specify that $\theta(e_G) = e_H$. On the other hand, most literature defines a ring homomorphism $h \colon R \to S$ with the law $h(1_R) = 1_S$.

This is because if for a monoid homomorphism $h \colon M \to N$,

$$h(e_M) = h(e_M^2) = h(e_M)^2.$$

If we assume $N$ admits inverses, multiplying by $h(e_M)^{-1}$ gives $h(e_M) = e_N$. But without any assumptions on inverses, we cannot derive that $h \colon e_M \mapsto e_N$.

If we instead assume that $N$ and $M$ are fields (we write $f \colon F_1 \to F_2$) with the laws $f(xy) = f(x)f(y)$ and $f(x+y) = f(x)+ f(y)$, we can consider the equation

$$f(1) = f(1^2) = f(1)^2.$$

So this boils down to solving the equation

$$a = a^2$$

for $a = f(1)$.

If $a = f(1) = 0$, then

$$f(x) = f(1x) = f(1) f(x) = 0 \cdot f(x) = 0$$

and we get a “trivial homomorphism”. One reason for blocking this solution might be because then the image $\operatorname{im} f = \{0\}$ is not a field (we cannot have $0 = 1$ in a field).

We can get the other solution to $a = a^2$ for a field by multiplying by $a^{-1}$, this gives us $a = 1$. This solution gives us exactly the axiom $f\left(1_{F_1}\right) = 1_{F_2}$.

Wikipedia formulates this phenomenon as the difference between “a semigroup homomorphism between monoids” and a ”monoid homomorphism”. Now: it is well-known that $\mathbf{Rng}$ is a “bad”/pathological category and a lot is known about the differences between $\mathbf{Rng}$ and $\mathbf{Ring}$.

But how different is $\mathbf{Ring}$ from the subcategory $\mathbf{Ring} \hookrightarrow \mathbf{Rng}$ with $\mathbf{Rng}$-morphisms? That is, what happens if we add all $\mathbf{Rng}$-morphisms “back” into $\mathbf{Ring}$?

Is it any "less" pathological than $\mathbf{Rng}$? It is any "friendlier" with respect to universal constructions, for example?

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    $\begingroup$ That's an interesting train of thought but "what happens" is vague. Do you have a more precise idea of what you're looking for? $\endgroup$ Commented Nov 2 at 17:14
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    $\begingroup$ @NajibIdrissi Yeah, this is an intentionally "soft" or "open" question, but I was looking to whether it is "less" pathological than $\mathbf{Rng}$ or "friendlier" with respect to universal constructions, for example. $\endgroup$ Commented Nov 2 at 17:19
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    $\begingroup$ Not really an answer, only a tangential comment. $\mathsf{Rng}$ has some niceties about it too: you get back inclusions into products $R \hookrightarrow R \times S$, ideals are not 'weird', but are actually subobjects (and so quotiens are between object and subobject) (this is actually quite illuminating, but is a long topic for a different time). Also, recall that adjoining a unit is generally not a lossless operation. $\endgroup$ Commented Nov 2 at 17:19
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    $\begingroup$ An observation along these lines: the forgetful functor from categories to nonunital categories is not fully faithful. But it is fully faithful when restricted to idempotent complete categories. $\endgroup$ Commented Nov 2 at 17:36
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    $\begingroup$ I don't think the linked post shows that there's anything pathological about $\mathbf{Rng}$ as a category, and indeed it's a perfectly nice category. Rather it shows that rngs have some behaviors that are unexpected, ring-theoretically. $\endgroup$ Commented Nov 3 at 19:58

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Since the question is somewhat vague, I will also provide a somewhat vague answer.

The question of how to deal with non-unital ring maps between actual rings actually appears in disguise in various places in topology. Under the usual "rings vs. affines" analogy, we can think of topological spaces as analogous to rings. For a continuous map $f : X \rightarrow Y$, the inverse image map $$ f^{-1} : \mathcal{O}(Y) \rightarrow \mathcal{O}(X)$$ behaves much like a ring homomorphism: It preserves arbitrary unions ( = sums ), and finite intersections ( = multiplication ). Dropping the requirement that $f^{-1}$ is unital is like not assuming that $f^{-1}(Y) = X$, which corresponds to considering partially defined continuous maps with open domain, where the open domain of definition is given by $f^{-1}(Y)$.

This should give you a feeling for how different the two categories are that you are considering. Of course, any partially defined continuous map factors as $$ X \supset f^{-1}(Y) \xrightarrow{f} Y$$ so the difference just amounts to understanding open inclusions (considered in the wrong direction) together with properly defined continuous maps.

Categorically, also some differences happen. For example the category $\mathrm{Ring} \hookrightarrow \mathrm{Rng}$ that you're considering is a pointed category: The $0$-ring is usually just a terminal object in the category of rings, but in the non-unital world it also becomes inital. Topologically, this corresponds to the fact that any space has a unique partially defined map to the empty set $\emptyset$. (it has empty domain of definition, of course).

Whether you want to work with non-unital ring homomorphisms / partially defined maps is up to you. There is an argument to be made that either notion can be reasonable. For example, the category of sets with partially defined functions appears naturally in many distinct categorical applications. (This category is of course just equivalent to the category of pointed sets - Here's where the analogy with non-unital rings becomes a bit more subtle)

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As a noncommutative ring-theorist, I wouldn't say that categorical considerations are ever the deciding factor of whether I work with unital vs. nonunital rings/homomorphisms. (Of course, for others it might.) Here are some of the aspects that have affected my work:

(Universal algebra): From a universal algebra perspective, homomorphisms should respect every operation in an algebra. A group is a universal algebra with operations $(\ast, (-)^{-1}, 1)$. So a group homomorphism should respect all three. We are somewhat spoiled by the fact that once a homomorphism respects the multiplication in a group, it automatically respects inversion and the identity.

Similarly, rings are universal algebras with operations $(+,\cdot, -, 0,1)$. So a ring homomorphism should respect all of these operations. The operations of negation and $0$ are automatically respected once addition is.

If we don't want to respect $1$, then we should be working with rngs, not rings, if the context is universal algebra.

(Simplicity): Checking that $I\subseteq R$ is an ideal is much easier in a unital ring, since closure under negation is a special consequence of absorption (just multiply by $-1\in R$).

Principal (left) ideals in a unital ring are of the form $Rx$, but in a nonunital ring they are a bit more difficult to properly write down; something like $\mathbb{Z}x+Rx$ or $R^{1}x$.

(Stronger results): In some situations you get stronger results by showing that a fact continues to hold in a unital ring. On the other hand, nonunital rings are the more general situation.

Thus, when proving abstract facts, one may generalize by working with nonunital rings if possible, or at least unital rings under nonunital homomorphisms. (If a property passes to corner rings, then unital rings under nonunital homomorphisms might be the right context, as mentioned by Benjamin Steinberg.) However, when constructing examples, it is best to work with unital rings and homs if possible, since unital rings are a special subclass of rngs.

(Associativity): Bjorn Poonen has an excellent article talking about how associativity of an operation really should require the existence of an identity for that operation. Since we are working in associative rings, we should require the presence of $1$.

(Point of origin): Rings most naturally arise as endomorphisms of additive structures. The identity map thus generally exists (and may need to be respected, if one doesn't wish to lose context).

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  • $\begingroup$ I don't think that associativity is a good reason to require the existence of a multiplicative unit. For example the ring of real valued continuous functions with compact support on, say, the real numbers, is associative, but doesn't have a multiplicative unit. A philosophy that excludes such natural objects seems to be very limited. $\endgroup$ Commented Nov 4 at 14:06

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