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).

  • 2
    $\begingroup$ A morphism $f: A \to B$ has a single source ($A$) and a single target ($B$). Hence the notation. $\endgroup$
    – Randall
    Apr 21, 2022 at 17:29
  • $\begingroup$ your Option B is correct, there isn't a standardized way to refer to $f$. $\endgroup$ Apr 21, 2022 at 17:44

1 Answer 1


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 $g(0) = 2$.

  • $\begingroup$ Just to elaborate, for all we know, we could have $f \in Hom(S_2, S_0)$. There need not be any connection between the “intrinsic nature” of $f$ and the role it plays in the category. $\endgroup$ Apr 21, 2022 at 17:57

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