Why is the limit the product, but not the coproduct?

On page 489 of Algebra, Paolo gives an example of a limit:

Let $I$ be the discrete category with 2 objects with only identity morphisms, let $\alpha$ be a functor from $I$ to any category $C$, and let $A_1=\alpha(1), A_2=\alpha(2)$, then the limit of $\alpha$ is the product of $A_1$ and $A_2$ in $C$.

My question is that, if the disjoint union of $A_1,A_2$ exists in $C$, then why can’t we take it to be the limit, with the morphisms required by the limit to be the selection of elements of certain indices in the disjoint union?

Solution 1:

Paolo is saying that by definition, this limit is the product. In principle, the limit could be naturally isomorphic to the coproduct (and indeed, in the case of Abelian categories like the category of Abelian groups, this is exactly what happens).

Let's recall that the definition of the limit of a diagram $F : I \to C$ is some object $X \in C$, together with a natural transformation $\pi : \Delta X \to F$, such that for any object $Y \in C$ and any natural transformations $\tau : \Delta Y \to F$, there exists a unique $f : Y \to X$ such that $\tau = \pi \circ \Delta f$. Here, $\Delta : C \to C^I$ is the diagonal functor.

Now, we should unpack this definition. A natural transformation $\pi : \Delta X \to \alpha$ consists of an arrow $\pi_1 : (\Delta X)(1) \to \alpha(1)$ and an arrow $\pi_2 : (\Delta X)(2) \to \alpha(2)$ - that is, an arrow $\pi_1 : X \to A_1$ and another arrow $\pi_2 : X \to A_2$. So the limit of $\alpha$ is an object $X \in C$, together with arrows $\pi_1 : X \to A_1$ and $\pi_2 : X \to A_2$, such that for all $Y \in C$, for all arrows $\tau_1 : Y \to A_1$ and $\tau_2 : Y \to A_2$, there is a unique $f : Y \to X$ such that $\tau_1 = f \circ \pi_1$ and $\tau_2 = f \circ \pi_2$.

But the above is exactly the definition of the product of $A_1$ and $A_2$.