Why is the image of a C*-Algebra complete? I am currently working through the book by Bratteli and Robinson on C* and W* algebras, there is one point at the beginning of chapter 2.3 that is frustrating me.
If we take *-morphism to be a function $\pi: U \to B$ between $C^*$-algebras $U,V$ with:


*

*$\pi$ is linear and multiplicative

*$\forall A \in U,$ $\pi(A^*)=\pi(A)^*$
Then a property is of such a map is $||\pi(A)||≤||A||$ (so contractive and as such continuous).
The book remarks that the image of $U$ is closed by "an easy" application of this continuity property. However I cannot reach the result and I feel like I am failing to do something extremely obvious. Does anybody have any tips?
(It is maybe interesting to note that in the erratum they comment that the line "by an easy" is to be omitted! One more comment: Since $U$ and $V$ are complete normed vector spaces, I am taking it to be equivalent for a subset to be complete or closed)
 A: To expand on my comment, let us take as given the following two (nontrivial) facts:

Lemma 1: If $U$ is a $C^*$-algebra and $I\subset U$ is a closed 2-sided $*$-ideal, then $U/I$ is a $C^*$-algebra.
Lemma 2: If $U$ and $V$ are $C^*$-algebras and $\pi:U\to V$ is an injective $*$-morphism, then $\|\pi(A)\|=\|A\|$ for all $A\in U$.

Now suppose $\pi:U\to V$ is any $*$-morphism of $C^*$-algebras; we wish to show $\pi(U)$ is closed.  Let $I$ be the kernel of $\pi$; then $\pi$ factors through an injective $*$-morphism $\pi':U/I\to V$.  By Lemma 1, $U/I$ is also a $C^*$-algebra, so we can apply Lemma 2 to $\pi'$ and conclude that $\pi'$ is an isometry.  In particular, this implies the image of $\pi'$ is closed.  But the image of $\pi'$ is the same as the image of $\pi$, hence the image of $\pi$ is closed.
Let me now briefly sketch the proofs of the lemmas.  In Lemma 1, it suffices to show that the usual quotient norm satisfies the $C^*$ identity.  This can be shown by some calculations using an approximate identity for $I$ (the idea being that the best way to approximate $A\in U$ by elements of $I$ is to take $AB_\alpha$, where $B_\alpha\in I$ is an approximate identity for $I$).   In Lemma 2, you can use the $C^*$ identity to reduce to the case that $A$ is self-adjoint, in which case the norms of $A$ and $\pi(A)$ can be computed as the spectral radius.  It thus suffices to show that $\pi$ preserves spectra of self-adjoint elements.  This can be shown using the continuous functional calculus (if $\sigma(A)$ were strictly larger than $\sigma(\pi(A))$, you could find two continuous functions on $\sigma(A)$ that agree on $\sigma(\pi(A))$, and this contradicts injectivity of $\pi$).
