Two-way Hash Functions
Yes, this is possible.  Here are two examples of such a function.

One function is $f(x)=x$.  This has no collisions, and the function is easy to invert.

Here is another function:  define $f(x) = x$ if $x$ is 256 bits long, otherwise $f(x) = \operatorname{SHA256}(x)$.  Then it is very hard to find collisions for $f$ (since SHA256 is collision-resistant).  At the same time, given $y$, it is easy to find an $x$ such that $f(x)=y$, so we might be willing to say that $f^{-1}(y)$ is easily computable.  To put it another way, given $y=f(x)$ where we are told $y$ but not told $x$, it is easy to find $x'$ such that $f(x')=y$.  The catch is that $x'$ might be different from the original $x$.