Let $ A , B $ be $C^{*}$-algebra with identity and $ \varphi : A \longrightarrow B $ is a linear map and for all $ a \in A \quad \parallel \varphi ( a ) \parallel = \parallel a \parallel $, so that $ \forall a \in A $, $ \varphi ( a^{*} ) = \varphi ( a )^{*} \qquad , \varphi ( 1_{A} ) = 1_{B} $.
prove:
$ \varphi ( A_{+} ) \subset B_{+} $
( $ A_{+}$ is the set of positive elements of $A$)
positive element means:
$$a=a^{*},~ \sigma (a) \subseteq [0,\infty) $$
If you means $\varphi$ is a $∗$-homomorphism. Then for any $a\in A_{+}$, then $\varphi (a)=\varphi ( \sqrt{a}^2 )=\varphi ( \sqrt{a})^2 $, so $\varphi (a)\in B_{+}$ (since it is a square). So $\varphi ( A_{+} ) \subset B_{+}$.
If you are confused with what is "$\sqrt{a}$", see here.