Closure of regular languages to shuffle using closure operations Given a language:
$L = \{\; a_1b_1a_2b_2a_3b_3\dots a_nb_n \mid \forall i: a_i,b_i \in \Sigma,  a_1\dots a_n \in L_1\ ,  b_1\dots b_n \in L_2 \;\}$

Also $L_1, L_2$ are regular languages.

Using closure only (homomorphism) prove that L is also regular language.



I think there can be a mapping $h\colon (L_1 \cup L_2) \to \Sigma$, then use $h^{-1}(\Sigma)$  in order to show regularity. I'm a bit stuck over here.
The idea is to create two copies of the alphabet $\Sigma$, say $\Sigma_1,\Sigma_2$; each $\sigma \in \Sigma$ corresponds to $\sigma^{(i)} \in \Sigma_i$. You now show that the following languages are regular:
$$
\begin{align*}
\tilde{L}_1 &amp;= \{a_1^{(1)}b_1^{(2)}\ldots a_n^{(1)}b_n^{(2)} : a_1 \ldots a_n \in L_1 \}, \\
\tilde{L}_2 &amp;= \{a_1^{(1)}b_1^{(2)}\ldots a_n^{(1)}b_n^{(2)} : b_1 \ldots b_n \in L_2 \},
\end{align*}
$$
where in both cases only some of the letters are constrained. Now take $\tilde{L}_1 \cap \tilde{L}_2$ and transform it to $L$.

I leave the rest of the details for you to complete.