Caveat in interpreting many-one reducibility as "computational difficulty"

Question

I am learning computability theory and recently was exposed to the following definition. For two sets $A, B \subseteq \mathbb N$, we say that $A$ is many-one reducible to $B$ and write $A \leq_m B$ if there exists a total computable function $f$ for which

$$ x \in A \iff f(x) \in B $$

on all $x \in \mathbb N$. The interpretation I was given for this is that $A \leq_m B$ means that computing membership in $A$ is "no harder than" computing membership in $B$. This makes some sense: if we treat total computable functions as being a priori "very easy to compute", then $A \leq_m B$ states that there is a "very easy" way to turn an indicator function $1_B$ for $B$ into one for $A$: let $1_A = 1_B \circ f$.

However, this interpretation of $\leq_m$ has a serious caveat; namely, that in general a set is not many-one reducible to its complement. For example, take the halting set $K = \{ n : \varphi_n(n) \downarrow \}$, and assume towards contradiction that all sets were many-one reducible to their complements. Then we'd have $\overline K \leq_m K$ (by assumption) and also $K \leq_m K$ (by reflexivity). Since a set $A$ is many-one reducible to $K$ exactly when it is c.e., then we'd know that $K$ and $\overline K$ are both c.e. And since if a set and its complement are both c.e. then that set is computable, we'd have that $K$ is computable, which is not true. Contradiction!

So it's not generally true that $A \leq_m \overline A$. But it is generally true that computing membership in $A$ is "no harder than" computing membership in $\overline A$: simply test for $a \in \overline A$ and negate the result.

My question, then, is this: is there a compelling interpretation for $\leq_m$ (as the flawed one is) without such serious defects? Alternatively, is there a compelling reason why we are interested in the ordering $\leq_m$ which requires our reduction $f$ to act on the input instead of the output? (I recognize the subjectivity of what is "compelling" so I have marked this as a soft question)

Answer 1

Quoting a comment from @NoahSchweber so I can close this question:

There are plenty of times where positive and negative information do behave differently; for example, it's easy to show that a number is in the halting problem, but it's hard to show that it isn't. $\leq_m$ preserves the "sidedness" at work here, which - sometimes - is something we care about. See also enumeration reducibility, which blends aspects of $\leq_T$ and $\leq_m$. –