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While studying the proof of Bott periodicity for operator $ K $-theory in this set of notes, I learned this fact:

Theorem. Let $ A $ and $ B $ be $ C^{*} $-algebras. Let $ f,g: A \to B $ be $ * $-homomorphisms. Then $ f + g $ is also a $ * $-homomorphism if and only if the ranges of $ f $ and $ g $ are orthogonal, i.e., $$ f[A] g[A] = g[A] f[A] = \{ 0_{B} \}. $$

In general, $ f + g $ is only a $ * $-preserving linear map, and multiplication may not be preserved unless further conditions are imposed. I managed to prove the theorem, but my argument is not entirely algebraic in the sense that it uses topological facts about $ C^{*} $-algebras.

My proof

The backward implication is trivial enough, so let us prove the forward one only.

Suppose that $ f + g $ is a $ * $-homomorphism. Then for all $ a_{1},a_{2} \in A $, we have \begin{align} (f + g)(a_{1} a_{2}) & = (f + g)(a_{1}) \cdot (f + g)(a_{2}) \\ & = [f(a_{1}) + g(a_{1})] [f(a_{2}) + g(a_{2})] \\ & = f(a_{1}) f(a_{2}) + f(a_{1}) g(a_{2}) + g(a_{1}) f(a_{2}) + g(a_{1}) g(a_{2}), \\ (f + g)(a_{1} a_{2}) & = f(a_{1} a_{2}) + g(a_{1} a_{2}) \\ & = f(a_{1}) f(a_{2}) + g(a_{1}) g(a_{2}). \end{align} It follows immediately that $ (\star) ~ f(a_{1}) g(a_{2}) + g(a_{1}) f(a_{2}) = 0_{B} $ for all $ a_{1},a_{2} \in A $.

Next, let $ a \in A $ be any self-adjoint element. As $ (\star) $ implies that $ f(a) g(a) = - g(a) f(a) $, we get $$ f(a^{2}) g(a^{2}) = f(a) f(a) g(a) g(a) = - f(a) g(a) f(a) g(a) = f(a) g(a) g(a) f(a), $$ and similarly, $$ g(a^{2}) f(a^{2}) = g(a) g(a) f(a) f(a) = - g(a) f(a) g(a) f(a) = f(a) g(a) g(a) f(a). $$ We also know from $ (\star) $ that $ f(a^{2}) g(a^{2}) + g(a^{2}) f(a^{2}) = 0_{B} $, so $ f(a) g(a) g(a) f(a) = 0_{B} $. Hence, by the self-adjointness of $ a $, we have $$ [f(a) g(a)] [f(a) g(a)]^{*} = f(a) g(a) g(a) f(a) = 0_{B}. $$ Therefore, $ f(a) g(a) = 0_{B} $, and by interchanging $ f $ and $ g $, we also obtain $ g(a) f(a) = 0_{B} $. As our choice of $ a $ was arbitrary, the discussion in this paragraph applies to all self-adjoint elements of $ A $.

Finally, let $ (e_{i})_{i \in I} $ be any self-adjoint approximate identity in $ A $. Then for all $ x,y \in A $, we get \begin{align} f(x) g(y) & = \lim_{i \in I} f(x e_{i}) g(e_{i} y) \qquad (\text{$ C^{*} $-homomorphisms are automatically continuous.}) \\ & = \lim_{i \in I} f(x) f(e_{i}) g(e_{i}) g(y) \\ & = \lim_{i \in I} f(x) ~ 0_{B} ~ g(y) \qquad (\text{By the previous paragraph.}) \\ & = 0_{B}. \end{align} Similarly, $ g(x) f(y) = 0_{B} $ for all $ x,y \in A $. This concludes the proof. $ \quad \blacksquare $


Question. Can we obtain the same result if we merely assume that $ A $ and $ B $ are $ * $-algebras over $ \Bbb{C} $? For convenience, we may suppose that $ (a^{*} a = 0_{A}) \Rightarrow (a = 0_{A}) $ for all $ a \in A $ and likewise for $ B $.

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Here is a small extension of your idea. You have, for any $a,b\in A $, $$ f (a)g (b)+g (a)f (b)=0. $$ Then $$ f (ab)g (ba)=f (a)f (b)g (b)g (a)=-f (a)g (b)f (b)g (a)=f (a)g (b)g (b)f( a) $$ and $$ g (ab)f (ba)=g (a)g (b)f (b)f (a)=-g (a)f (b)g (b)f (a)=f (a)g (b)g (b)f (a). $$ Now $$ 0=f (ab)g (ba)+g (ab)f (ba)=2f (a)g (b)g (b)f (a). $$ When $a,b $ are selfadjoint we get $$ f (a)g (b)[f (a)g (b)]^*=f (a)g (b)g (b)f (a)=0, $$ and we conclude that $f (a)g (b)=0$ for all selfadjoint $a,b $. But then, as any $x,y\in A $ can be written $x=a+ib $, $y=c+id $, $$ f (x)g (y)=f (a+ib)g (c+id)=f (a)g (c)-f (b)g (d)+i [f (b)g (c)+f (a)g (d)]=0. $$

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  • $\begingroup$ This is awesome, Martin. Thanks! :) With this, we have the result that if $ A $ and $ B $ are $ * $-algebras over $ \Bbb{C} $, and $ b^{*} b = 0_{B} \Rightarrow b = 0_{B} $ for every $ b \in B $, then the sum of any two $ * $-homomorphisms $ f,g: A \to B $ is yet again a $ * $-homomorphism if and only if the ranges of $ f $ and $ g $ are orthogonal. $\endgroup$ – Transcendental Jul 22 '15 at 15:37

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