0
$\begingroup$

Consider the Hom-Set Functor $H : C^{op} \times C \rightarrow \mathbf{Set}$.

If we let $c_1, c_2$ be categories (and assume, as is tradition, that all hom-sets are actually sets), then we have that $H(c_1, c_2)$ is a hom-set in $\mathbf{Set}$.

Question: Since $h \in H(c_1, c_2)$ is a function from $Ob(c_1)$ to $Ob(c_2)$, doesn't that imply that $Ob(c_1)$ and $Ob(c_2)$ are themselves sets? But how can we assume that?

  • 1
    The assumptions of this question aren't clear. It sounds as if you're assuming local smallness of $C$, and that $C$ includes, as objects, some categories. But we know nothing about what morphisms we might find between $c_1,c_2$, and different answers to this may make for very different answers to your question. If the hom-sets on $C$ really are collections of functions between particular sets, you've implicitly made the assumption that $Ob(c_n)$ (for $n=1,2$) is a set when you assumed local smallness of $C$. – 2017-02-16

1 Answers 1

1

Let me try to lay things out straight.

  • Given a category $\mathbf{C}$ with small hom-sets, there's a functor $$\mathrm{Hom} : \mathbf{C}^{op} \times \mathbf{C} \rightarrow \mathbf{Set}.$$

  • Given objects $X$ and $Y$ in $\mathbf{C}$, $\mathrm{Hom}(X,Y)$ is the set of all morphisms $X \rightarrow Y$.

  • The elements of $\mathrm{Hom}(X,Y)$ aren't necessarily functions.

  • Supposing $\mathbf{C} = \mathbf{Set}$, then $X$ and $Y$ are sets and $\mathrm{Hom}(X,Y)$ does indeed consist of functions. But the domain of $f \in \mathrm{Hom}(X,Y)$ is $X$, and not $\mathrm{Ob}(c_1)$ for some auxilliary category $c_1$.

Does that help?

  • 0
    Isn't the domain of $f$ in this context $Ob(X)$, strictly speaking? – 2017-02-16
  • 0
    @user1770201 - Why would it be? – 2017-02-16
  • 0
    @user1770201, in the usual way of writing mathematics, we don't write $\mathrm{Ob}(X)$ when $X$ is a set. We just write $X$, and we usually refer to its elements as elements, not objects. – 2017-02-16
  • 0
    The domain of the *object part* of the $\mathrm{Hom}$ functor is $\mathrm{Ob}(\bf C)\times\mathrm{Ob}(\bf C)$, and it can indeed be a proper class. – 2017-02-16