Non-invertible Karp reduction Karp (many-one) reducibility between $NP$-complete problems $A$ and $B$ is defined as a polynomial-time computable function $f$ such that $a \in A$ if and only if $f(a) \in B$. Berman-Hartmanis isomorphism conjecture states that between any pair of $NP$-complete problems there is bijection $f$ which is computable and invertible in polynomial-time.
I am interested in polynomial-time invertability of Karp reductions between natural NP-complete problems. Every Karp reduction $f$ is polynomial-time computable but it is not clear whether $f^{-1}$ is also computable in polynomial-time.
I am looking for Karp reduction $f$ between two natural NP-complete problems where its inverse $f^{-1}$ is not known to be polynomial-time computable.

Is there a pair of natural NP-complete problems A, B  and polynomial-time computable injective reduction $f$ from A to B (where $f^{-1}$ is not know to be computable in polynomial-time)?

P.S. Natural problem means that the problem is not an artificially made up problem to answer the question (or similar ones) and people are interested in the problem independently (defined by Kaveh).
 A: Such a non-invertible reduction exists between any pair of “natural” NP-complete problems (and conjecturally, between any pair of NP-complete problems), as long as non-invertible poly-time functions exist at all.
While the Berman–Hartmanis conjecture in general is open, various particular cases of it hold unconditionally. In particular, “natural” NP-complete problems are known to satisfy the conjecture. So, let me simply define for the sake of this post that an NP-complete language is natural if it is poly-time isomorphic to SAT.
This means all natural NP-complete languages are paddable, and in particular, any such language $A$ is poly-time isomorphic to the language $A\oplus\emptyset=\{0_\smile x:x\in A\}$. Now, if $A$ and $B$ are two such languages, let $f$ be any reduction of $A$ to $B$ (which may be asumed to be a poly-time isomorphism if desired), and let $g\colon\{0,1\}^*\to\{0,1\}^*$ be an arbitrary poly-time function. Then the function $f\oplus g$ defined by
$$\begin{align}
(f\oplus g)(0_\smile x)&=0_\smile f(x),\\
(f\oplus g)(1_\smile x)&=1_\smile g(x),
\end{align}$$
is a reduction of $A\oplus\emptyset$ to $B\oplus\emptyset$, which retains all bad properties of $g$; so, if we chose $g$ not poly-time invertible, then $f\oplus g$ is not poly-time invertible either. We can construct a reduction of $A$ to $B$ with the same properties by composing with the poly-time isomorphisms $A\simeq_p A\oplus\emptyset$ and $B\oplus\emptyset\simeq_p B$.
