What really is a morphism in a category?

I have had quite some exposure with category theory this year. I have even completed a quite long course in category theory and did very well in it. So I thought I am quite good at it. But, something very striking happens with this subject; something that I haven't experienced before with any other area of mathematics I attempt to learn.

It seems like the more I learn about category theory, the more unconfident I feel about what I really know.

For some reason, I had not stumbled upon the fact that $\mathbf{Set}$ is not equivalent to its opposite - until now I didn't really care to understand what actually IS the opposite category. Unfortunately I have realized that I have a gap in my understanding of this idea.

By trying to trace back where the gap is, I have arrived to the conclusion that I do not really, honestly, morally, understand what a morphism is.

What do we really mean with the notation $f:A\to B$ in a category? Is a morphism $f:A\to B$ dependent on the objects $A$ and $B$ or not? If yes, in what way and how much? If not, what do $A$ and $B$ stand for, given $f$ - and are they explicit? For example, in the category $\mathbf{Set}$, the phrase "add" should be some sort of morphism between two sets of "numbers". Is this A morphism in the category of sets, just by itself? If yes, how can you say if its domain and codomain are the sets, e.g of natural numbers rather than the sets of rational numbers? And how do you discriminate between them? Or do you have to specify two objects of the category first (sets of numbers here) and then describe explicitly how the morphism "acts" on them (assigns to the numbers their sum)?

I believe what would really help is to hear how people actually realize and think about the notion of a morphism. Which I guess is a question I should have asked myself a long time ago.

• Maybe we can think of a morphism as a function between objects, satisfying some conditions. – Sigur May 9 '14 at 17:04
• But this is the whole point - that you can not and in fact you should not think of a morphism as a function. – Justanothernerd May 9 '14 at 20:45

You can think of a category as a directed graph with extra structure (namely identity arrows and composition). From this perspective, objects are the vertices of the graph and morphisms are the (directed) edges, where $f : A \to B$ means the edge is directed from $A$ to $B$. Taking the opposite category means reversing the direction of all of the edges. (Exercise: construct a directed graph, not yet a category, which fails to be isomorphic, as a directed graph, to its opposite.)

"Add" could be interpreted as, for example, a morphism $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, where you've picked out the particular object $\mathbb{R}$.

• But, what is the morphism in the "add" case? Is "add", as a process, the morphism and the source and target just some sort of "representatives" we choose to demonstrate what it does, or is the morphism dependent in some specific way to the source and target objects? It feels like the source and target are a bit mysterious as notions. – Justanothernerd May 9 '14 at 20:55
• @Justanothernerd: What is addition? If you want, addition is addition among the reals, $\Bbb R\times\Bbb R\to\Bbb R$. As you said, there is also an addition among the rationals, $\Bbb Q\times\Bbb Q\to\Bbb Q$. Actually, any Abelian group $A$ admits an 'addition', by very definition, which is a morphism $A\times A\to A$ in $\Bbb{Set}$. – Berci May 9 '14 at 21:52
• @Justanothernerd: "add" is a bad example to think about here; there are additional complications depending on what exactly you mean by "add." If I'm only interested in addition of real numbers, then "add two real numbers" refers to a specific morphism in $\text{Set}$, namely the function $+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ given by $(a, b) \mapsto a + b$. What you seem to be confused about is the categorical status of "the platonic ideal of addition," divorced from any particular kind of things you want to add. This can also be addressed in the language of category theory... – Qiaochu Yuan May 9 '14 at 21:56
• ...as follows. Suppose we decide that addition means the group operation on a group (it will be misleading here to restrict our attention to abelian groups). Then I claim that "addition" refers to the following thing: if $U : \text{Grp} \to \text{Set}$ is the forgetful functor from groups to sets, then "addition" is a natural transformation $U \times U \to U$. This description neatly packages the group operation on all groups simultaneously. See qchu.wordpress.com/2013/06/09/operations-and-lawvere-theories for some details. – Qiaochu Yuan May 9 '14 at 21:57
• (The reason I didn't want to restrict attention to abelian groups is that confusingly, for an abelian group $A$ addition is actually a morphism $A \times A \to A$ of abelian groups, which is false in the nonabelian case.) – Qiaochu Yuan May 10 '14 at 4:28

As far as I know, according to any definition you might have of a category, given any arrow $f\in Arr(\mathscr{C})$, where $\mathscr{C}$ is any category, there exists a unique pair of objects $(A,B)$ of $\mathscr{C}$ such that $f\in\hom_{\mathscr{C}}(A,B)$ and indeed, whenever $(A,B)\neq (A',B')$, $\hom_{\mathscr{C}}(A,B)\cap \hom_{\mathscr{C}}(A',B')=\emptyset$. Thus, $f\colon A\to B$ precisely means that $f\in \hom_{\mathscr{C}}(A,B)$ for a unique pair of objects $(A,B)$ of $\mathscr{C}$, and that is what $A$ and $B$ stand for.

Hope this helps somehow.