Showing isomorphism of two $C^*$ algebras It seems that quite a standard trick of showing two $C^*$ algebras are as follows: 

Let $A$ be a $C^*$ algebra $B$ another $C^*$ algebra. $A' \subseteq A$ be a subalgebra that is closed under $*$. (Not necessarily closed) and is dense in $A$. Let $L:A' \rightarrow B$ be a $*$-isometric homomoprhism which has dense image.  Then $L$ extends to a $*$-isometric isomorphism .

So my aim is showing that $L$ actually extends to an injective map - since  it is not necessarily case that $L$ is injective in its closure. 
 A: Given $a\in A$ take $\{a_j\}\subset A'$ with $a_j\to a$. Since $L$ is isometric, we have 
$$
\|L(a_j)-L(a_k)\|=\|L(a_j-a_k)\|=\|a_j-a_k\|,
$$
so $\{L(a_j)\}$ is Cauchy. As $B$ is complete, there exists $b_a=\lim L(a_j)$. We want to extend $L$ by defining $L(a)=b_a$. For this we need to show that $b_a$ is unique for each $a$; and this follows easily: if $c_j\to a$, then 
$$
\|L(c_j)-L(a_j)\|=\|c_j-a_j\|\leq\|c_j-a\|+\|a-a_j\|\to0, 
$$
so $L(c_j)\to b_a$. So the extension $L$ is well-defined. Also
$$
\|L(a)\|=\|\lim_j L(a_j)\|=\lim_j\|L(a_j)\|=\lim_j\|a_j\|=\|a\|,
$$
so $L$ is isometric. 
If you don't require $L$ to be isometric, the result is not true. Let $A=C[0,1]$, $B=\mathbb C\oplus C[0,1]$, and take $A'\subset A$ to be the $*$-algebra of complex polynomials. Define $L:A'\to B$ by 
$$
L(p)=p(2)\oplus p. 
$$
This is clearly an injective $*$-homomorphism. And it has dense image: Given $(\lambda,f)\in B$, we can choose a sequence of polynomials $\{p_j\}$ such that $p_j\to f$ uniformly on $[0,1]$ and $p_j(2)=\lambda $ (simply use Stone-Weierstrass on a continuous function $g$ that agrees with $f$ on $[0,1]$ and with $g(2)=\lambda$). 
So $L$ satisfies all the hypotheses; but it is not bounded. If you consider $p_n(x)=x^n$, then $\|p_n\|=1$, while $p_n(2)=2^n$, so $\|L(p_n)\|=2^n$. 
A: The answer by Martin is of course correct, I just want to emphasise that this statement and the way to see it is essentially elementary functional analysis. In my opinion is very helpful to be able to differentiate when a statement is about the linear structure and when a statement needs the C*.
Central is the following statement: Let $A:V\to W$ be a continuous linear map between normed vector spaces, then there is a unique continuous linear extension $\overline A :\overline V\to\overline W$ between the completions. This can be done as an elementary exercise and is basically the first half of Martin's answer.
(As a remark: The above statement is itself an expression of a more general statement, namely that Lipschitz maps between metric spaces $A,B$ so that $B$ is complete and $A$ is dense in some other metric space $A'$ always have a unique continuous extension $A'\to B$.)
The rest of what you need follows from:


*

*If a continous linear map is isometric on a dense sub-set, then it is isometric on its entire domain.

*If a continuous linear map is multiplicative wrt continuous multiplications on a dense sub-set, then it is multiplicative on its entire domain.

*If a continuous linear map $A$ satisfies $A(v^*)= A(v)^*$ on a dense sub-set and the $*$ operation is continuous on domain and codomain then the above equality holds on the entire domain.
Note that this is basically the same statement $3$ times: Continuous linear maps that preserve a continuous structure on a dense sub-set automatically do so on the entire domain.
