Regarding Banach algebra homomorphism

Question

Let $A$ be a Banach algebra with identity $e_A$ and let $B$ be a Banach algebra with identity $e_B$. Let $\phi:A\longrightarrow B$ be a Banach algebra homomorphism i.e $\phi$ is linear and $\phi(ab)=\phi(a)\phi(b)$. Then is $\phi(e_A)=e_B?$

Answer 1

The answer to the question exactly as you asked it is no, because it could be that $\phi(x)=0$ for every $x$.

Edit: In the original I said that was the only counterexample. Thanks to Omno-something for pointing out that that's nonsense. A less trivial counterexample: Define $\phi:C([0,1])\to C([0,1]\cup[2,3])$ by $$\phi f(x)=\begin{cases}f(x),&(x\in[0,1]), \\0,&(x\in[2,3]).\end{cases}$$

In fact

$\phi(e_A)=e_B$ if and only if there exists $x\in A$ such that $\phi(x)$ is invertible.

One direction is trivial, since $\phi_B$ is invertible.

If there exists $x$ with $\phi(x)$ is invertible then $$e_B\phi(x)=\phi(x)=\phi(e_Ax)=\phi(e_A)\phi(x),$$which implies $\phi(e_A)=e_B$ since $\phi(x)$ is invertible.

Answer 2

Not necessarily. For instance, with $A = \Bbb C^{n \times n}$ and $B = \Bbb C^{m \times m}$ with $n < m$, we find that $$ \phi(M) = \pmatrix{M&0\\0&0} $$ is a suitable map for which $\phi(e_A) \neq e_B$.