Let $A$ and $B$ be C*-algebra, $h\colon A\rightarrow B$ is *-homomorphism.
If $a\in A_{\operatorname{sa}}$, then $\operatorname{sp}(h(a))\backslash \{0\}\subset \operatorname{sp}(a)\backslash\{0\}$. Here, $A_{\operatorname{sa}}$ denotes all the self adjoint elements in $A$
How to prove this inclusion?
This result holds in more general case of Banach algebras.
By $A_+:=A\oplus_1 \mathbb{C}$ we denote unitalization of Banach algebra $A$. Multiplication in $A_+$ is given by the formula $$ (a,\lambda)(b,\mu ) =(ab+\lambda b+\mu a,\lambda\mu) $$ The algebra $A_+$ is unital with unit $e_{A_+}=(0_A,1_{\mathbb{C}})$.
For a given Banach algebra homomorphism $h:A\to B:a\mapsto h(a)$ we have well defined unitalization $h_+:A_+\to B_+:(a,\lambda)\mapsto (h(a),\lambda)$. Clearly $h_+$ is a unital homomorphism, i.e. preserve units. We also use the notation $rg_A(a)=\mathbb{C}\setminus sp_A(a)$
Lemma 1. For a given unital Banach algebra $A$ we have $sp_{A_+}(a)=sp_A(a)\cup\{0\}$ for all $a\in A$.
Proof. Let $\lambda\in\mathbb{C}\setminus\{0\}$. Let $b\in A$ be inverse element of $a-\lambda e_A$, then it is easy to check that $b-\lambda^{-1}(e_{A_+}-e_A)\in A_+$ is the inverse element of $a-\lambda e_{A_+}$. Conversely, if $c\in A_+$ is the inverse element of $a-\lambda e_{A_+}$, then $e_Ac\in A$ is the inverse element of $a-\lambda e_A$.
From previous paragraph it follows that $rg_A(a)\setminus \{0\}=rg_{A_+}(a)$, hence $sp_{A_+}(a)=sp_A(a)\cup\{0\}$.
Lemma 2. Let $h:A\to B$ be a unital homomorphism of unital Banach algebras, then $sp_B(h(a))\subset sp_A(a)$ for all $a\in A$
Proof. Let $\lambda\in\mathbb{C}$. Assume $b\in A$ is the inverse element of $a-\lambda e_A$. It is easy to check that $h(b)\in B$ is the inverse element of $h(a)-\lambda e_B$. This means that $rg_A(a)\subset rg_{B}(h(a))$, hence $sp_B(h(a))\subset sp_A(a)$.
Proposition. Let $h:A\to B$ be a (not necessarily unital!) homomorphism of unital Banach algebras, then $sp_B(h(a))\cup\{0\}\subset sp_A(a)\cup\{0\}$.
Proof. Consider unital homomorphism $h_+:A_+\to B_+$. By lemma 2 we have $sp_{B_+}(h_+(a))\subset sp_{A_+}(a)$ for all $a\in A_+$. Recall that $h_+(a)=ha(a)$ for all $a\in A$, so by lemma 1 we get that $sp_B(h(a))\cup\{0\}\subset sp_A(a)\cup\{0\}$.
