Let E1 and E2 be complex vector spaces, and let u1 and u2 be semilinear involutions on E1 and E2 respectively. Let f be a linear mapping of E1 into E2. Then f∘u1=u2∘f if and only if f(E1u1)⊂E2u2.
If f∘u1=u2∘f, we obtain immediately f(E1u1)⊂E2u2. Conversely, suppose that this condition is satisfied. Let x∈ E. Write x=x1+ix2 with (x1,x2)∈E1u1×E1u1 (lemma above). We then have f(u1(x))=f(x1)−if(x2) and u2(f(x))=u2(f(x1))−iu2(f(x2))=f(x1)−if(x2)=f(u1(x)).
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.