Abstract index is an invariant of Banach algebras

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I just started to read the book An introduction to Operator Algebras by Kehe Zhu. In this book the author defines the index group of a Banach algebra $A$ (which is assumed to be a unital algebra) as the quotient space $I(A)=G(A)/G_0(A)$; here $G(A)$ is the set of all the invertible elements in $A$ and $G_0(A)$ is the connected component of $1$. Next, the autor claims that if $\phi: A \longrightarrow B$ is a homomorphism between two Banach algebras $A$ and $B$ then $\phi$ induces a group homomorphism between $I(A)$ and $I(B)$. Moreover, it is claimed that if $A$ and $B$ are isomorphic then so are $I(A)$ and $I(B)$. However, it seems that some other hypothesis are needed. For example, the natural way of inducing an homomorphism between $I(A)$ and $I(B)$ is by defining $f:I(A) \longrightarrow I(B)$, $xG_0(A) \mapsto \phi(x) G_0(B)$. Of course, one have to check that this is well defined in which case we need to see that if $xG_0(A)=yG_0(A)$ then $\phi(x)G_0(B)=\phi(y)G_0(B)$. The equation $xG_0(A)=yG_0(A)$ implies that $xy^{-1} \in G_0(A)$ and therefore $\phi(x)\phi(y^{-1}) \in \phi(G_{0}(A))$. Thus, if $\phi$ is continuous then $\phi(G_{0}(A))$ is a connected set containing 1 (also assuming that $\phi(1)=1$) and therefore $\phi(G_{0}(A)) \subset G_0(B)$. Hence, $\phi(x)\phi(y^{-1}) \in G_0(B)$ and therefore $\phi(x)G_0(B)=\phi(y)G_0(B)$. However, as you can see, I assumed that $\phi$ is a continuous map and also $\phi(1)=1$ but the book never mentioned these hypothesis. So, are my suppositions superfluous? Otherwise, how can we possibly prove the assertions?

In advance thank you very much.