Are regular languages 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:

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