I have asked the following question on M.SE [here](https://math.stackexchange.com/questions/2147151/showing-using-exponentials-that-phi-is-a-jordan-morphism?__=1163220907), but I have not yet received a response. 
I do apologize of this is not the correct site to post it on - if so, please do let me know and I will remove the question if it is off-topic.


I am reading through a research paper (*Spectrum preservig linear mappings in Banach algerbas* by *B. Aupetit and H. du T. Mouton*) in which they say the following:



> 
> If $\phi$ is a linear mapping from a Banach algebra $A$ into another one $B$ such that $\phi(1)=1$ and $\phi(x)^{-1}=\phi(x^{-1})$ for $x$ invertible, then using exponentials it is easy to prove that $\phi$ is a Jordan morphism.
> 
> 
> 


How does one show the above statement true using exponentials?


My initial thought, is defining another mapping $\psi$ as some kind of "combination" of $\phi$ and the exponential, i.e. something in the line of 


$$\psi(x) = \phi(\exp(x))$$ or $$\psi(x)=\exp(\phi(x))$$ or every $x \in A$.


Is this kind of reasoning correct? And, if so, can someone help guide me on the correct path? I am not looking for a complete proof of the result - only a little "nudge" in the right direction :).



Showing that $\phi$ is a Jordan morphism
A standard trick one sees in Banach algebra theory, relating additive and multiplicative structure, is to look at $\exp(\lambda a)$ for fixed $a\in A$ as a holomorphic function of $\lambda$, and then try to play games with power-series or Liouville's theorem or other complex-analytic techniques.


The given conditions on $\phi$ imply that $\phi(x)\phi(x^{-1})=1\_B$ for all $x\in A$. In particular
$$\phi(e^{\lambda a})\phi(e^{-\lambda a}) = 1\_B$$


The LHS is a $B$-valued holomorphic function of $\lambda$, and we can work out the first few terms in the power-series expansion:
$$
\begin{aligned}
\phi(e^{\lambda a})\phi(e^{-\lambda a}) & = 
\phi( 1\_A+ \lambda a + \frac{\lambda^2}{2}a^2 + \dots) \phi(1\_A - \lambda a + \frac{\lambda^2}{2}a^2 + \dots ) \\
& = (\phi(1\_A)+ \lambda\phi(a) + \frac{\lambda^2}{2} \phi(a^2) + \dots) (\phi(1\_A) - \lambda\phi(a) + \frac{\lambda^2}{2} \phi(a^2) + \dots) \\
& = 1\_B +\lambda^2 (\phi(a^2) - \phi(a)^2 ) + O(\lambda^3)
\end{aligned} 
$$


From this we obtain $\phi(a^2)=\phi(a)^2$ for all $a\in A$, which is the Jordan homomorphism condition.


Actually, now that I have worked through this, one doesn't seem to need the full power of being holomorphic. You could use $\lambda\in{\bf R}$ with $\lambda\to 0$ to achieve the same effect.


