Proof regular languages are closed under homeomorphism

Question

Let $\Sigma_1 , \Sigma_2$ be alphabets. Let $L\subseteq \Sigma_1^*$ be a regular language, and let $ h:\Sigma_1^* \rightarrow \Sigma_2^* $ be a homomorphism.

Proof $h(L)$ is regular.

I have written a proof using induction over the number of operands in the regular expression $r$ such that $L = L(r)$, that there exists a regular expression $r'$ such that $L(r') = h(L)$.

I divided it to three cases:

1. $r=r_1 \cup r_2$

2. $ r= r_1r_2$

3. $ r =r_1^*$

Since functions union might not always be a function (i.e $f(1)=1$, $g(1) =2$ $\rightarrow f\cup g$ not a function) I would like to get your opinion for the union case, to verify I didn't do anything wrong there - especially in lines $5,6,7$:

1. $w\in h(L) \Leftrightarrow w = h(\sigma_1 \dots \sigma_n)\in h(L) \wedge \sigma_1 \dots \sigma_n \in L \Leftrightarrow $
2. $\sigma_1 \dots \sigma_n \in L(r_1 \cup r_2) \Leftrightarrow $
3. $\sigma_1 \dots \sigma_n \in L(r_1) \cup L(r_2) \Leftrightarrow $
4. $\sigma_1 \dots \sigma_n \in L(r_1) \vee \sigma_1 \dots \sigma_n\in L(r_2) \Leftrightarrow $
5. $h(\sigma_1 \dots \sigma_n) \in h(L(r_1)) \vee h(\sigma_1 \dots \sigma_n) \in h(L(r_2)) \Leftrightarrow $
6. $h(\sigma_1 \dots \sigma_n) \in h(L(r_1)) \cup h(L(r_2)) \Leftrightarrow $

Now, using the induction hypothesis, since $r_1$ and $r_2$ has less operands,

7. $h(\sigma_1 \dots \sigma_n) \in L(r_1') \cup L(r_2') \Leftrightarrow $
8. $w \in L(r_1' \cup r_2') $

Is this a valid proof for this case?

Answer 1

This proof looks correct. You can shorten it a bit (it doesn't need to be so descriptive), and you would get the following proof:

$h(L)=h(L(r_1\cup r_2))=h(L(r_1)\cup L(r_2)) = h(L(r_1))\cup h(L(r_2)) = L(r'_1)\cup L(r'_2)=L(r'_1\cup r'_2)$

Where the transition $h(L(r_1)\cup L(r_2)) = h(L(r_1)) \cup h(L(r_2))$ can be shown for any function $h$ (not necessarily a homeomorphism!).