If both the concatenation of two languages and the second "half" are regular, is the first too? Given that $L_2$ is regular and infinite and $L_1 \cdot L_2$ is regular, then $L_1$ is also regular.


I need some help on getting started on proving this is the case.

My intuition is that if $L_1 \cdot L_2$ is regular there exists a DFA for it, in which case you can just remove the states that correspond to the $L_2$ part of the DFA and you are left with a DFA that recognizes $L_1$.
Suppose that $L_2 = \Sigma^*$ and $L_1$ includes the empty string. Then $L_1 \cdot L_2 = \Sigma^*$, yet $L_1$ doesn't even have to be computable.