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Proof. If $f$ is universally closed, then $B$ is integral over $A$ by Lemma 29.44.7. Hence $B'$ is integral over $A$ and $f'$ is universally closed (by the same lemma). This proves the case where $f$ is universally closed.

Continuing, we see that $B$ is integral over $B'$ (Algebra, Lemma 10.36.15) which implies $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ is surjective (Algebra, Lemma 10.36.17). Thus if $A \to B$ induces purely inseparable extensions of residue fields, then the same is true for $A \to B'$. This proves the case where $f$ is universally closed and universally injective, see Lemma 29.10.2.

The case where $f$ is a universal homeomorphism follows from the remarks above, Lemma 29.45.5, and the obvious observation that if $f$ is surjective, then so is $f'$.

If $A \to B$ induces isomorphisms on residue fields, then so does $A \to B'$ (see argument in second paragraph). In this way we see that the lemma holds in the remaining case. $\square$