# Why NP-Complete reduction is not reversible?

I have read the question asked here Is polynomial reduction reversible and the logic actually makes sense to me. In other words, if A is polynomially reducible to B, it means that A <= B in terms of hardness.

However, when I thought more about the reduction process, I am confused. As far as I can understand, the NP-Complete reduction process is as follows:

(1) Find some NP-Complete problem (i.e. A), convert its inputs and output to the inputs and output of the problem we want to prove (i.e. B)

(2) Show that if A is YES, then B is YES

(3) Show that if B is YES, then A is YES.

Then I also read several classic reduction proofs, and here is my confusing part. In all the reductions I read, step (1) seems to be a "one-to-one mapping" between the inputs of A to the inputs B. For example, minimum set cover (MSC) to minimum vertex cover (MVC), edges in MVC is elements in MSC and nodes in MVC is sets in MSC. Since it's a "one-to-one mapping" of inputs A to inputs B, if we can convert the inputs of A to the inputs of B, then we can always convert the inputs of B to the inputs of A. In this case, isn't step (1) a reversible process? I am not sure if "one-to-one mapping" is the correct terminology, but the main idea I want to say is "the specific way" of mapping inputs of A to the inputs of B. Is there any reduction example that maps the input from A to the inputs of B, but the mapping is not reversible, i.e. can't map the inputs of B back to A?

What's more, no matter if we convert A to B or convert B to A, I think step (2) and step (3) above are still the same. Therefore, I am really confused about which part makes the NP-Complete reduction not reversible?

Here is a concrete example. Let $$A$$ be the problem in which all instances are NO instances, and let $$B$$ be the problem with a unique YES instance $$1$$. The function $$f(x) = 0$$ is a reduction from $$A$$ to $$B$$, but there is no reduction from $$B$$ to $$A$$, since there is nothing we can map $$1$$ into.