Are regular languages closed under inverse homomorphism?

Let $\Sigma$ and $\Delta$ be alphabets. Consider a
function $\varphi: \Sigma \rightarrow \Delta^*$. Extend $\varphi$ to
a function from $\Sigma^ \rightarrow \Delta^$ such that:
\begin{eqnarray*}
\varphi(\varepsilon) & = & \varepsilon \\
\varphi(w\sigma ) & = & \varphi(w)\varphi(\sigma ), \textrm{ for any }
w \in \Sigma ^*, \sigma \in \Sigma
\end{eqnarray*}
Any function $\varphi:\Sigma^ \rightarrow \Delta^$ defined in this way
from a function $\varphi: \Sigma \rightarrow \Delta^*$ is called a
homomorphism. We can extend this definition to languages as follows:
for any language $L$ and homomorphism $\varphi$, let
$$\varphi(L)=\{\varphi(w) : w\in L\}.$$

Are regular languages are closed under inverse homomorphism?

Yes, regular languages are closed under inverse homomorphism.

Here is a proof.
We start with a DFA of $L$, $X$. To prove that $\varphi^{-1}(L)$ is regular, we will construct a DFA, $Y$ for $\varphi^{-1}(L)$.

$$ X = (Q,\Sigma,\delta_X,q_0,F)$$

So our construction will be as follows:

$$ Y = (Q,\Delta,\delta_Y,q_0,F)$$

Where:

$$\delta_Y(q,w) = \hat{\delta_X}(q,\varphi(w))$$

Or in other words, the transition in Y on w is the result of the sequences of transition that X makes on the string of symbols $\varphi(w)$. So all we need to do now is prove this transition. To do this we will induct on $|w|$. Our hypothesis being $\hat{\delta_Y}(q,w) = \hat{\delta_X}(q,\varphi(w))$.

Base: When $w = \epsilon$:

$$q_0 = \delta_Y(q_0,\epsilon) = \delta_X(q_0,\varphi(\epsilon)) = \delta_X(q_0,\epsilon) = q_0$$

Inductive Step: First let $w = \sigma s$:

$$\hat{\delta_Y}(q_0,w) = \hat{\delta_Y}(\delta_Y(q_0,\sigma),s)$$

Then by our inductive hypothesis:

$$\hat{\delta_Y}(\delta_Y(q_0,\sigma),s) = \hat{\delta_Y}(\delta_X(q_0,\varphi(\sigma)),s)$$

Now by our DFA definition above:

$$\hat{\delta_Y}(\delta_X(q_0,\varphi(\sigma)),s) = \hat{\delta_X}(\delta_X(q_0,\varphi(\sigma)),\varphi(s))$$

Pulling the $\delta$ back in:

$$\hat{\delta_X}(\delta_X(q_0,\varphi(\sigma)),\varphi(s)) = \hat{\delta_X}(q_0,\varphi(\sigma)\varphi(s))$$

Now by the definition of homomorphism:

$$\hat{\delta_X}(q_0,\varphi(\sigma)\varphi(s)) = \hat{\delta_X}(q_0,\varphi(\sigma s)) = \hat{\delta_X}(q_0,\varphi(w))$$

StackExchange Computer Science Q#14785, answer score: 9