# If $A\leq_m^\mathsf{}B$ and $B$ is a regular language, does that imply that $A$ is a regular language?

Corollary:1 We know that if $$A\leq_m^\mathsf{}B$$ and $$B$$ is decidable then $$A$$ is also decidable.

This is because if there exists a specific algorithm for solving $$B$$ and we can also reduce $$A$$ to $$B$$ then we can have a solution of $$A$$ as well. Hence $$A$$ is decidable.

We know that regular languages is always decidable.

Corollary:2 So from (Corollary:1)if $$A\leq_m^\mathsf{}B$$ and $$B$$ is regular then $$A$$ is should also be regular.

But I have made this wrong my taking Counterexample. For example, define the languages $$A = \{ 0^n1^n \mid n\ge 0 \}$$ and $$B = \{1\}$$ both over the alphabet $$\sum = \{0, 1\}.$$ Define the function $$f : \sum^*\to \sum^*$$ as $$f(x) := \begin{cases} 1,& \text{if}\ w\in A,\\0,&\text{if}\ w\not\in A\end{cases}$$ Observe that $$A$$ is a context-free language, so it is also Turing-decidable. Thus, $$f$$ is a computable function. Also, $$w \in A$$ if and only if $$f(w) = 1,$$ which is true if and only if $$f(w)\in B.$$ Hence,$$A\leq_m^\mathsf{}B.$$ Language $$A$$ is nonregular, but $$B$$ is regular since it is finite.

Above example proving that $$A$$ isn't necessarily regular. But according to Corollary:1, $$A$$ must be regular. I don't understand where I did make mistake during understanding of theory.

Corollary $$2$$ is false.
It seems that you are using the implication "$$L$$ is regular $$\implies$$ $$L$$ is decidable" in the wrong direction. If $$A \le_m B$$ and $$B$$ is regular, Corollary 1 only tells you that $$A$$ is decidable, which does not imply that $$A$$ is regular.