Proving that Sp$\phi(x) \subset $ Sp$x$ for every $x \in A$ is equivalent to saying that $\phi$ preserves invertibility. Let $A$ and $B$ be unital Banach algebras with $\phi:A \to B$ a linear mapping.
I am looking specifically at linear mappings which preserves invertiblity and noticed that different papers use different definitions to define preserves invertibility. The one that comes up most, is the following:

Sp$\phi(x) \subset $ Sp$x$ for every $x \in A$.

I am trying to see why this is true though, i.e. why this is equivalent to $x$ invertible $\implies \phi(x)$ invertible.
Suppose that $x \in A$ is invertible and that $\phi$ preserves invertibility with $\phi(x^{-1})=\phi(x)^{-1}$.
Let $\lambda \in $ Sp$\phi(x)$, then $\phi(x) - \lambda\textbf{1}_B$ is not invertible in $B$.
I need to show that $\lambda \in $ Sp$x$. But I am not sure how to do so.
I have thought of the following:
Since $\phi(x) - \lambda\textbf{1}_B$ is not invertible in $B$ we have, in particular, that $\phi(x)\neq \lambda$. But I am not sure if this will take me anywhere.
Can anyone please point me in the right direction?
 A: In proofs such as this, it is easier to work with the contrapositive.
If $x-\lambda$ is invertible, then $\phi(x-\lambda)=\phi(x)-\lambda$ is invertible.  Thus $\mathbb C\setminus\sigma(x)\subset\mathbb C\setminus\sigma(\phi(x))$, and therefore $\sigma(\phi(x))\subset\sigma(x)$.
On the other hand, if $\sigma(\phi(x))\subset\sigma(x)$ for all $x\in A$, then supposing $x\in A$ is invertible means $0\notin\sigma(x)$. Thus $0\notin\sigma(\phi(x))$, and so $\phi(x)$ is invertible.
A: Suppose that $\phi$ is a unital linear mapping $\phi \colon A \to B$, which preserves invertibility. In particular, you have that if $\phi(x)$ is not invertible then it follows that $x$ is not invertible.
Let $\lambda \in \mathrm{Sp}(\phi(x))$. Then
$$
\phi(x) - \lambda \;1_B = \phi(x - \lambda \; 1_A)
$$
is not invertible. So $x - \lambda \; 1_A$ is not invertible and thus $\lambda \in \mathrm{Sp}(x)$.
For the converse assume that $\mathrm{Sp}(\phi(x)) \subseteq \mathrm{Sp}(x)$. Then, if $\phi(x)$ is not invertible, $0 \in \mathrm{Sp}(\phi(x)) \subseteq \mathrm{Sp}(x)$ and thus $x$ is not invertible.
