Computing colimit of $\bullet \rightarrow \bullet$ in Set?

Question

$\newcommand{\cSet}{\mathsf{Set}} \newcommand{\colim}{\operatorname{colim}}$ I'm trying to learn how to compute colimits in $\cSet$, so I'm testing out simple examples. Let's try a diagram $J \equiv \bullet \rightarrow \bullet$. Le $F: J \to \cSet$ be a functor whose image looks like $A \xrightarrow{f} B$. If I'm following nLab correctly, then the colimit is computed as $\colim F \equiv A \sqcup B / \sim$ where $\sim$ is the equivalence relation generated by the equations $a \sim f(a)$ for all $a \in A$.

As far as I can tell, this $\colim F$ is isomorphic to $B$ --- we have $a \in A$, which gets glued onto some $f(a) \in B$. So, every $a \in A$ is "annihilated" by being glued to a $f(a) \in B$. These don't even impose new relations, since each $a \in A$ gets mapped to a single $f(a) \in B$. So we must have $(A \cup B / \sim) \simeq B$?

I'm a little worried by this since the dual situation in the case of limits gives us $\lim F \equiv A \times B$ which is "more than" the colimit which seems to be equal to $B$.

Questions

1. Is the assertion that $\colim F \simeq B$ correct?
2. Is there some way to explain the "discrepancy" between the limit (being non-trivial) and the colimit (being trivial)?

Answer 1

The limit of $F$ is $A$, while the colimit of $F$ is $B$.

This is actually true for any functor $F : J \to C$ mapping to $A \to B$.

We first prove that the limit is $A$. For consider the maps $1_A : A \to A$ and $f : A \to B$. Suppose we have a third object $C$ and maps $g : C \to A$, $h : C \to B$ such that $h = f \circ g$. Then the unique map is $g : C \to A$. We see that $1_A \circ g = f$ and $f \circ g = h$.

To prove the same is true for the colimit, we simply take the limit in the opposite category.

Answer 2

$\newcommand{\cSet}{\mathsf{cSet}}$ To clarify, in terms of the general construction of limits in $\cSet$, we must compute the limit of the diagram $A \xrightarrow{f} B$ by constructing the set $\lim F \equiv \{ (a, b) : f(a) = b \}$. This set is isomorphic to $A$, since there is a unique b for each $a$ such that $f(a) = b$.