Let $A$ be a $*$-algebra, i.e. an algebra $A$ together with a map $*: A \to A$ such that
$$(a+ \lambda b)^* = a^* + \overline{\lambda} b^*$$ $$a^{**}= a$$ $$(ab)^* = b^* a^*$$
My book then claims that if $I$ is an ideal of $A$ satisfying $I=I^*$, then $A/I$ becomes a $*$-algebra for:
$$(a+I)^*= a^* + I$$
I proved this but I can't see where the hypothesis that $I=I^*$ is used. Can this hypothesis be ommited?
It can't be omitted, because without the assumption $I=I^*$ the involution $(a+I)^*=a^*+I$ is not well defined. Every time when we define a function on equivalence classes which does something with the representatives we have to prove the definition doesn't depend on the choice of the representatives.
So here, suppose $a+I=b+I$ and we have to prove that in that case $(a+I)^*=(b+I)^*$, i.e we have to show that $a^*+I=b^*+I$. And this is indeed true, because: ($a+I=b+I$ implies $a-b\in I$)
$a^*-b^*=(a-b)^*\in I^*=I$
So we used the assumption $I^*=I$ here.
Without this assumption the involution would not be well defined, because we would have some $a\in I$ such that $a^*\notin I$. In that case we would have $a+I=0+I$ but $(a+I)^*=a^*+I\ne 0+I=(0+I)^*$.