I'm feeling quite a bit embarrassed to ask such a basic thing, but I can't seem to figure it out. Let $R$ be a commutative ring and $A$ be a commutative $R$-algebra. Is the fork $$ A \xrightarrow{i} A \otimes_R A \rightrightarrows A \otimes_R A $$ an equalizer, where $i(a) = a \otimes 1$ and the two parallel arrows are given by $f(a \otimes b) = ab \otimes 1$ and $g = id$?

Yes, it is. Let a tensor $t$ be in the equalizer of $f$ and $g$. Then, $f\left(t\right)=g\left(t\right)$. If we write $t$ in the form $\sum\limits_{j\in I} a_j\otimes b_j$ (with $I$ being a finite set, and $a_j$ and $b_j$ being elements of $A$), then this rewrites as $\sum\limits_{j\in I} a_jb_j\otimes 1 = \sum\limits_{j\in I} a_j\otimes b_j$. Thus, $t = \sum\limits_{j\in I} a_j \otimes b_j = \sum\limits_{j\in I} a_j b_j \otimes 1 = i\left(\sum\limits_{j\in I} a_j b_j\right) \in i\left(A\right)$. Thus, the equalizer is contained in $i\left(A\right)$. The reverse inclusion is even more trivial, so the equalizer is actually equal to $i\left(A\right)$.