Proving that $\{0^i10^i : i \ge 1\}$ is non-regular, using only closure results I have been stumped on the following question for a few hours now, I feel like I am missing some "aha" moment.


  $\text{Suppose that } \{ a^nb^n : n \ge 1 \} \text{ is non-regular.}$
  $\text{Prove using closure results that } \{ 0^i10^i : i \ge 1 \} \text{ is non-regular.}$


Starting off I assumed for contradiction that the language is regular. Then I took countlessly many compliments and intersections and homomorphisms, none leading me anywhere close. 

I understand using Pumping Lemma would quickly solve this problem, but the question restricts the proof technique to not using Pumping Lemma.

How would one go about solving such a question with the given restraints? Is there a methodology as to find the correct closure results to use or is it mostly intuition and luck?
I would say that luck and intuition are improved by experience :)

The trick is also to use inverse morphisms! They can perform nondeterministic letter substitutions.

A related example.
Consider the morphism $h: \{a,b\}^*\to \{0\}^*$  with $h(a) = h(b) = 0$, then $h^{-1}$ maps a word $0^n$ to any string $w\in\{a,b\}^*$ with length $n$.  

Background. It is easy to write a FSA with output that maps strings $0^i10^j$ to strings $a^ib^j$, and regular languages are closed under finite state transductions (input-output transformation by two-tape finite state automaton). If you are not familiar with this result, by Nivat's Theorem: any finite state transduction can be written as a composition of an inverse morphism, intersection with regular language, and a morphism.