I am not sure to correctly understand the notion of morphism in category theory. To try to better understand, let's take a very simple example. Let's say that we have a category $\mathcal{C}$:
- whose objects are three singleton sets of natural numbers $S_{0} = \{0\}, S_{1} = \{1\}, S_{2} = \{2\}$
- whose morphisms are $f: \forall x \in S, f\left(x\right) \rightarrow x + 1$ and $g: \forall x \in S, g\left(x\right) \rightarrow x + 2$
When specified in that way, I am trying to understand which option is the correct one:
- Option A: $f$ is a morphism
- or Option B: $f$ is just a nice way to call two different morphisms: $f_{0}$ (whose source object is $S_{0}$ and target object is $S_{1}$) and $f_{1}$ (whose source object is $S_{1}$ and target object is $S{2}$) (in that case what is the correct mathematical notion corresponding to $f$ since $f_{0}$ and $f_{1}$ are the morphisms ?)
But that triggers another question. If the correct option is A, then I guess $\mathcal{C}$ is not a category because if we apply $f$ to $S_{2}$ the codomain ($S_{3} = \{3\}$) is not in $\mathcal{C}$ as it should be. If the correct option is B then if I understand correctly every morphism is specific to a single source object and to a single target object in the category.
So which option is the correct one? A clarification (with simple illustrative examples if necessary) would be very welcomed.
Bonus question: If the correct option is B, is there a way to call a category that would be "closed under its families of morphisms (if we call $f$ and $g$ families of morphisms), meaning all the domains and codomains for all possible compositions of morphisms would be in $\mathcal{C}$" (it's probably very handwavy but I hope you'll get what I mean).
To start with, you did not define a category, since there should be an identity morphism for each object. Next, neither A nor B is correct. You need to define precisely, for each pair of objects $(S_i, S_j)$ the morphisms from $S_i$ to $S_j$. If I understand your idea, you want to define $f_0: S_0 \to S_1$ by $f_0(0) = 1$, $f_1: S_1 \to S_2$ by $f_1(1) = 2$ and $g = f_1 \circ f_0: S_0 \to S_2$ by $g(0) = 2$.