Reduction between two decision problems
First, your definition of reducibility needs to include some sort of restriction on the power of&nbsp;$f$: typically, for example, we require the function to be computable.

Without this restriction, any two decision problems (except the trivial problems "always yes" and "always no") are reducible to each other.  Since there are always $y$ and&nbsp;$n$ such that $Q_2(y)=1$ and $Q_2(n)=0$, there is always a function&nbsp;$f$ that reduces $Q_1$ to&nbsp;$Q_2$: just take
\begin{equation*}
  f(x) = \begin{cases}
         \ y &amp;\text{if }Q_1(x)=1\\
         \ n &amp;\text{if }Q_1(x)=0\,.
         \end{cases}
\end{equation*}

So, we typically restrict to computable functions&nbsp;$f$ as reductions.  Your question is, essentially, is a computable reduction always reversible? The answer to this is no and the reduction&nbsp;$f$ above gives a big hint why: that function is not invertible and there's no reason, in general, to expect that even a computable reduction should be invertible.

For example, consider any decidable problem&nbsp;$Q_1$, and let $Q_2$ be the halting problem.  In this case, we can take $y$&nbsp;to be the description of any Turing machine that halts immediately, and let $n$&nbsp;be any Turing machine that just moves the head right forever.  Since $Q_1$&nbsp;is decidable, the reduction&nbsp;$f$ above is computable.  It tells you how to translate questions about&nbsp;$Q_1$ to questions about the halting problem.  It also kind of tells you how to translate questions about whether or not $y$ and&nbsp;$n$ halt into questions about&nbsp;$Q_1$.  But it doesn't help you determine whether any other Turing machine halts&nbsp;&ndash; and it can't, because that problem is undecidable.