Add a line to sketch of proof

THanks to Tom Graber
https://stacks.math.columbia.edu/tag/03R1#comment-5066

CONTRIBUTORS

@@ -114,6 +114,7 @@ Alberto Gioia
 Charles Godfrey
 Julia Ramos Gonzalez
 Jean-Pierre Gourdot
+Tom Graber
 Matt Grimes
 Darij Grinberg
 Jonathan Gruner

etale-cohomology.tex

@@ -9089,7 +9089,13 @@ \section{Brauer groups}
 \end{theorem}
 
 \begin{proof}[Sketch of proof]
-In the abelian case ($d = 1$), one has the identification
+To prove that $\delta$ defines a group homomorphism, i.e., that
+$\delta(A \otimes_K B) = \delta(A) + \delta(B)$, one computes
+directly with cocycles.
+
+\medskip\noindent
+Injectivity of $\delta$. In the abelian case ($d = 1$), one has the
+identification
 $$
 H^1(\text{Gal}(K^{sep}/K), \text{GL}_d(K^{sep})) =
 H_\etale^1(\Spec(K), \text{GL}_d(\mathcal{O}))
@@ -9099,7 +9105,10 @@ \section{Brauer groups}
 \cite{SGA4.5}.) Rather, to prove this, one can reinterpret $\delta([A])$ as the
 obstruction to the existence of a $K$-vector space $V$ with a left $A$-module
 structure and such that $\dim_K V = \deg A$. In the case where $V$ exists, one
-has $A \cong \text{End}_K(V)$. For surjectivity, pick a
+has $A \cong \text{End}_K(V)$.
+
+\medskip\noindent
+For surjectivity, pick a
 cohomology class $\xi \in H^2(\text{Gal}(K^{sep}/K), (K^{sep})^*)$,
 then there exists a finite Galois extension $K \subset K' \subset K^{sep}$
 such that $\xi$ is the image of some