Jordan-homomorphism; equivalent properties Let $\phi:A\to B$ a linear map between $C^*$-algebras $A$ and $B$. I want to know, why the following properties are equivalent: $(i) \phi(ab+ba)=\phi(a)\phi(b)+\phi(b)\phi(a)$ and $(ii) \phi(a^2)=\phi(a)^2$ for all $a,b\in A$.
The direction => is ok: Let $a\in A$, it is $\phi(aa+aa)=\phi(2a^2)=\phi(a)\phi(a)+\phi(a)\phi(a)=2\phi(a^2)$. Now, $\phi$ is linear, therefore if you multiply both sides with $\frac{1}{2}$, it is $\frac{1}{2}\phi(2a^2)=\phi(\frac{1}{2}*2a^2)=\phi(a^2)=\frac{1}{2}2\phi(a)^2=\phi(a)^2$. You obtain $(ii)$.
How to prove the other direction <=? I started with $\phi(aa)=\phi(a)\phi(a)$ for every $a\in A$. Therefore it is $\phi(aa+bb)=\phi(aa)+\phi(bb)=\phi(a)\phi(a)+\phi(b)\phi(b)$. Now I'm stuck. Does $\phi(aa)=\phi(a)\phi(a)$ for all $a\in A$ imply $\phi(ab)=\phi(a)\phi(b)$ for all $a,b\in A$? If not, how to continue? Regards

If we assume that $\varphi(a^2)=\varphi(a)^2$ for all $a$, then for $a,b$, we have, using linearity and the square-preserving property,
$\varphi((a+b))^2=(\varphi(a)+\varphi(b))^2=\varphi(a)^2+\varphi(a)\varphi(b)+\varphi(b)\varphi(a)+\varphi(b)^2.$
Now, observe that
$\varphi((a+b)^2)=\varphi(a^2+ab+ba+b^2)=\varphi(a^2)+\varphi(ab+ba)+\varphi(b^2)$.
By assumption, these two equations are the same, and we know that squares are preserved.  Then comparing the equations gives $\varphi(ab+ba)=\varphi(a)\varphi(b)+\varphi(b)\varphi(a)$.
If I recall correctly, a Jordan homomorphism need not be a homomorphism in the usual sense.
