If a is a positive element then it has a unique positive square root, i.e. a unique b positive such that b^2=a.
I understand the existence part of the proof. If we define $f(x)=\sqrt x$ then $f\in C(\sigma (a))$ as $\sigma (a)$ non-negative. $b=f(a)$ satisfies $b^2=a$ and $b\ge0$.
My advisor gave me another proof for the uniqueness part and I want to understand it, in addition to the "popular" proof in which we use the fact that $f\circ g (a) = f(g(a))$ for $f,g\in C(\sigma (a))$ (continuous functional calculus respects composition).
So, consider the special case of a C*-algebra of the form C(X). Positive elements here are continuous functions with nonnegative values. Here we can directly see that there is only one positive square root.
In general, suppose $c$ is some other positive square root of $a$. We want to show that $c = f(a)$.
$a$ and $f(a)$ are elements in the C*-subalgebra $C^*(1,c)$, which is commutative. But $C^*(1,c)$ is isomorphic to $C(X)$ for some compact Hausdorff $X$, so we're back to the previous paragraph.
Now, I understand the proof, but I'm a little bit worried that we "identify" between positive elements in isomorphic $C^*$ algebras. Is it true that *-isomorphism preserves positive elements? do we really need this fact to complete the proof?
Thank you.
Once you know that positive elements are those of the form $a^*a$, you have $$ \pi(a^*a)=\pi(a^*)\pi(a)=\pi(a)^*\pi(a)\geq0. $$ This shows that any $*$-homomorphism preserves positivity.
If you restrict to isomorphisms, you can show it directly also if you define "positive" as "selfadjoint and $\sigma(a)\subset[0,\infty)$". Because $a-\lambda I$ is invertible if and only if $\pi(a)-\lambda I=\pi(a-\lambda I)$ is invertible.