What Is a Morphism?

Question

I know that there are several posts on the same question. They all ask for examples for morphisms that are not functions. So, morphisms are more general than functions; they are the arrows connecting the objects of a category. However, I still cannot avoid the idea that they are functions.

By function, I mean exactly that the relation is well-defined. If $(a, b_1), (a, b_2)$ are inside the graph of $f: A \to B$, then $b_1 = b_2$. Therefore, if a morphism, say $g: A \to B$, is not a function, then I am allowed to have $(a, b_1), (a, b_2)$ inside the graph of $g$.

But, without the condition that they are well-defined, I do not see how associativity holds. For example, let:

1. The graph of $f: A \to B$ contains $(a, b)$;
2. The graph of $g: B \to C$ contains $(b, c_1), (b, c_2)$;
3. The graph of $h: C \to D$ contains $(c_1, d_1), (c_2, d_2)$.
4. The graph of $f': A \to D$ contains $(a, d_1), (a, d_2)$

So, $h((g \circ f)(a)) = d_1$ or $d_2$? Similarly, $(h \circ g)(f(a)) = d_1$ or $d_2$? This is what I mean; if I am forced to choose either $(a, d_1)$ or $(a, d_2)$, then indeed I am dealing with functions ....

Answer 1

First, you're making too many assumptions. Why should morphisms be relations? Why should they have a graph? Why should A and B have elements?

Second, even with these assumptions, morphisms still don't need to be functions. Consider the category of Sets with arrows reversed. Then each morphism is like the reverse of a function. In particular, there are 2 distinct morphisms from {x,y} to {x} even though there is only one such function.

Answer 2

Morphisms, as already pointed out, could not be functions, or relations at all. In my category algebra class we saw morphisms as "arrows" between objects of the category. If we are in a concrete cathegory, as in the cathegory of sets or groups, a morphism is reasonably a function, a homomorphism, or something like this. However in an abstract category, such as in a category with two natural numbers $n, m$ as objects and morphisms are given by $n\times m$ matrices, relations don't play any role. You still want to assure that the morphisms are componible.