While studying the proof of Bott periodicity for operator $ K $-theory in [this set of notes](http://perimeterinstitute.ca/personal/tfritz/2010/bott_periodicity.pdf), 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 $.
> 
> 
> 



Does the following $ C^{*} $-algebraic result have a purely algebraic proof?
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 $,
$$
f (x)g (y)=f (a+ib)g (c+id)=f (a)g (c)-f (b)g (d)+i [f (b)g (c)+f (a)g (d)]=0.
$$


