Can two morphisms with equal mappings be distinct in category theory?

Question

I'm trying to get an understanding of equality in category theory, and found some answers to this question on SE (here and here), but I didn't really feel like I fully grasped the notion of equality from those answers so I thought I'd ask about equality in a more specific context.

Given two objects $A$ and $B$ in some category $C$, we can introduce a morphism $A \rightarrow B$ and assign it a label such as $f$ and write that label assignment as $f: A \rightarrow B$. If another morphism $g$ would be introduced as $g : A \rightarrow B$ we would now have two labels for morphisms from $A$ to $B$.

My question is, are the morphisms $f$ and $g$ introduced here meaningfully distinct? Or are they exactly equal (in the same sense that writing $x=5$ and $y=5$ implies $x=y$) even in a general category where we're not modelling some other mathematics?

If they are not exactly equal, how can it be possible to formalize things such as monomorphisms and epimorphisms that seem to rely on some notion of equality of morphisms?

Answer 1

Recall that part of the definition of a category is that for any two objects $A$ and $B$, the category specifies a set (or class, if your categories are not required to be locally small) $\operatorname{Mor}_C(A,B)$ of morphisms from $A$ to $B$. When we write $f:A\to B$, that is just an alternate notation for saying that $f\in\operatorname{Mor}_C(A,B)$. So if $f:A\to B$ and $g:A\to B$, do $f$ and $g$ have to be equal? Certainly not, because $\operatorname{Mor}_C(A,B)$ is just some set, and $f$ and $g$ might be different elements of that set.

In other words, writing $f:A\to B$ does not specify exactly what $f$ is in the same way that writing $x=5$ specifies exactly what $x$ is. All it says is that $f$ is some element of the set $\operatorname{Mor}_C(A,B)$, but it doesn't say which one. This is analogous to stating something like "$x$ is a real number"--this tells you some information about $x$, but does not uniquely determine it. If you say something like "let $x$ and $y$ be real numbers", then $x$ and $y$ might be equal, or they might not--you simply haven't said whether they are or not. Similarly, if you say "let $f:A\to B$ and $g:A\to B$", this says nothing about whether $f$ and $g$ are equal.

For a really simple example, consider the category $C=\mathtt{Set}$ of sets, in which objects are sets and morphisms are functions between sets. If you have two sets $A$ and $B$, are any two functions $A\to B$ the same? Certainly not in general. For instance, if $A=B=\mathbb{R}$, you might have one function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$ and another function $g:\mathbb{R}\to\mathbb{R}$ defined by $g(x)=x^2$.

Answer 2

There are various ways to address your question.

From a naive point of view, it is equivalent to: "someone gave me a set. Does the set have more than two elements?" The answer is: sometimes yes, but oftentimes it can be very difficult to know it does.

More explicitly: in the axioms of a category, there is one that says more or less

for evey two objects $X,Y$ of $\cal C$ there is a set $\hom(X,Y)$ such that...

Sets have elements; $f : A\to B$ is just another way to say that $f \in \hom(X,Y)$. Now, if $f,g \in \hom(X,Y)$ it is perfectly legitimate to ask if $f=g$ (and depending on your foundation of reference the question can have several different flavours). The answer may vary; let $\mathcal C$ be the category $$ 0 \overset{u}\to 1\underset{g}{\overset{f}\rightrightarrows} 2 $$ defined by setting $fu=gu$ and $f=g$ if and only if the Riemann hypothesis is true (or an equally mainstream open problem in hard mathematics).

Of course, some mathematicians may reject this "definition" as a figment; the point is that at least in its naive definition, a category knows how to distinguish its parallel morphisms at least as much as set theory knows how to distinguish elements of a conglomerate.

A different point of view is that of enriched category theory: an enriched category is not a collection of objects and sets of arrows, but instead it is a collection of objects and other objects ${\cal C}(X,Y)\in \cal V$ of a monoidal category such that composition and identities... yadda yadda, you can read the definition on WikiPedia or on the $n$Lab.

Now being an arrow $f :X \to Y$ oftentimes means that $f$ can be thought as the element of a set. But this results in a loss of information, and it is not considered a good practice when doing enriched CT.

In this setting, an enriched category knows how to distinguish its parallel morphisms at least as much as the category $\cal V$ where its hom-objects live... well, now you see the problem: a set has elements. An object of $\cal V$ does not, and in general it doesn't consist of the sum of irreducible parts.