$f+g$ $*$-homomorphsim if and only if $im \, f \cdot im\, g = 0$

Question

Let $f,g:A \rightarrow B$ be $*$-homomorphisms of $C^*$ algebras. Then $f+g$ is a $*$-homomorphism if and only if $im \, f \cdot im \, g =0$.

How does this hold?

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My thoughts: We know that $f+g$ is linear. Repsects $*$ since $$(f+g)(a^*) = f(a)^*+g(a)^*= (f+g)(a)^*$$

We have $$(f+g)(ab) = f(ab)+g(ab) = f(a)f(b)+g(a)g(b) \quad (1)$$ $$ (f+g)(a) \cdot (f+g)(b) = f(a)f(b)+ f(a)g(b)+g(a)f(b)+g(a)g(b) \quad (2)$$

So

$f+g$ is $*$ - homomorphism iff $$ f(a)g(b)=-g(a)f(b)$$ for all $a,b \in A$.

Hence $\Leftarrow $ is clear. But $\Rightarrow$ not so, we can at most deduce $f(a)g(a)=0$.

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This is stated in page 4, proof of Lemma 7.

Answer 1

Let $\{p_j\}$ be an approximate unit for $A$. As you said, we have $f(p_j)g(p_j)=0$ for all $j$. Now, for any $a,b\in A$, \begin{align} f(a)g(b)=\lim_j f(ap_j)g(p_jb)=\lim_jf(a)f(p_j)g(p_j)g(b)=0. \end{align}