Context-free languages are closed under union, concatenation, Kleene star, and homomorphism

Rename variables of $G_1$ and $G_2$ apart, and add a fresh start variable $S$. The grammar with productions of $G_1$ and $G_2$ together with $S\to S_1$ and $S\to S_2$ generates exactly $L_1\cup L_2$. Similarly, adding only the bridge rule $S\to S_1{S}_2$ yields a grammar for the concatenation $L_1{L}_2$.

For Kleene star, keep $G_1$ and add a fresh start variable $S$ with productions $S\to \epsilon$ and $S\to S_{1}S$. Repeating the second rule any finite number of times and then using $S\to \epsilon$ generates exactly a concatenation of finitely many words from $L_1$, so the new grammar generates $L_1^{\ast }$.

For homomorphism, first rename the variables of the chosen grammar for $L_1$ so that they are disjoint from $\mathrm{\Delta }$. Then replace every terminal letter $a$ occurring in a production by the word $h(a)$, keeping that renamed variable set unchanged. The result is still a context-free grammar over $\mathrm{\Delta }$, and if a derivation in the source grammar yields $w=a_0\cdots a_{n-1}$, then the corresponding derivation in the rewritten grammar yields $h(a_0)\cdots h(a_{n-1})=\widehat{h}(w)$ by [L2].

Conversely, any terminal derivation in the rewritten grammar follows the same variable-expansion pattern as a terminal derivation in the renamed source grammar, because the only change was to replace each source terminal occurrence by its prescribed image word $h(a)$. Therefore the terminal word produced at the end is exactly the image under $\widehat{h}$ of a word of $L_1$, so the new grammar generates exactly $\widehat{h}(L_1)$.

Steps 1.1, 1.2, 1.3, and 2.1 construct grammars for all four languages, so [L1] shows that context-free languages are closed under union, concatenation, Kleene star, and homomorphism.