Probably a stupid question but, need help with clarification of condition for equality of two maps.

Question

I'm reading Conceptual Mathematics: An First Introduction to Categories by Lawvere and Schanuel and in it, they provide a condition to test for the equality of two maps, here's the link.

Here's what I (think I) know. Roughly, a category is a collection of objects and arrows between the objects such that:

1) for every object, there is an identity arrow

2) for each pair of arrows where the domain of the second arrow is the codomain of the first arrow, there is a composite arrow

3) the arrows must satisfy the identity and associativity laws.

So, based on the image which describes the condition to test for the equality of two maps, what if the category doesn't have singleton sets? As in, what if the category is just the objects $A$ and $B$, with arrows $1_A, 1_B, A \xrightarrow{f} B, \text{and } A \xrightarrow{g} B$? How can I test to see if $f = g$ if there are no third singleton sets in the category? Sorry if my question doesn't make any sense, I'm just not really understanding. Thank you!

Answer 1

The extract states (in bold no less) that the condition $(\forall a : 1 \to A.f \circ a = g \circ a) \implies f = g$ is a "test for equality of maps of sets". That is, in the category of sets and functions, you can characterize equality by the above test. And it's easy to see this fairly directly corresponds to the normal notion of function extensionality. This definition is not being provided as a general definition of equality of morphisms in any category for the reasons you state (among others). Most likely it's a lead in to the fullness and faithfulness of the Yoneda embedding which does give a general condition applicable and true of all categories. In particular, $f = g : A \to B$ if and only if $\forall X\in\text{Ob}(\mathcal{C}).\forall x : X \to A.f \circ x = g \circ x$.