In Conway's A Course in Operator Theory, proposition 1.7 (e) is
If $\rho:A \to B$ is $\ast$-homomorphism, then $\|\rho(a)\| \leq \|a\|$ for all $a$ in $A$.
and the beginning of the proof is
First note that by adjoining an identity if necessary, it can be assumed that $A$ is unital. Now, the definition of a $\ast$-homomorphism does not assume that $\rho(1)$ is the identity of $B$. How ever it is easy to see that $\rho(1)$ is an identity for $\rho(A)$. So there is no loss of generality in assuming that $B$ has an identity and $\rho(1) = 1$.
I could understand "$\rho(1)$ is identity for $\rho(A)$", but I'm stack at "there is no loss of generality in assuming that $B$ has an identity and $\rho(1) = 1$".
Please some advice to understand this. Thank you for your time.
A $*$-homomorphism $f: A \to B$ is such that $f(a)^* = f(a^*)$.
Hence, we have $\rho(1)^* = \rho(1^*) = \rho(1)$ since $1^* = 1$. Moreover, $\rho(1)$ behaves similarly to $1_B := 1 \in B$. Thus if $\rho(1)$ has this key defining property similar to $1_B$ then there is no fault in assuming WLOG $\rho(1) = 1_B$.