I'm reading Lenstra's notes on the étale fundamental group, and I've got stuck on his exercise 3.9(c). He says that if $A$ and $B$ are separable algebras over a field $K$, and $f:A \to B$ a $K$-algebra homomorphism, then the image of $f$ is precisely $\{b \in B: b \otimes 1 = 1 \otimes b\text{ in }B \otimes_A B\}$.
By his theorem 2.7, every separable $K$-algebra is a finite product of separable extension fields $K_i$ of $K$. So the morphisms of two such algebras are fairly restricted: if I'm not mistaken, if $A = \prod K_i$ and $B = \prod L_j$, and $v_i$ and $w_j$ are the respective idempotents defining these splittings, then each $f(v_i)$ is a sum of distinct $w_j$, each generating an extension field of $K_i$.
I don't know how to get from this to the claim. More generally, for any map of commutative rings $A \to B$, we have this subring $\{b\in B:b \otimes 1 = 1 \otimes b \text{ in }B \otimes_A B\}$, which I feel like I've seen in other contexts as well, though I can't remember where. Is there a high-level way of thinking about this set, or more general conditions that tell you it's equal to $f(A)$? Thank you in advance.