Do injection preserves inequalities?

Question

Suppose $f$ is injective; is it true that

$$ \Pr[f(x)<y] \qquad \text{is equivalent to} \qquad \Pr[x<f^{-1}(y)]? $$ provided $y\in\mathrm{Ran} f$? What if $f(x)=-x$?

Answer 1

No! Injectivity is a statement about distinct inputs having distinct outputs.

This does not contain any content about inequalities: if your claim was true, then it would be one of those "surprisingly deep" things. But it's absolutely false. As you say, $f:x\mapsto -x$ is a counterexample. Any decreasing injective function is a counter-example.

But now you've edited the question to be about probability, and without defining your probability space, your probability distributions, etc. the question can't really be answered. I'd still say "no" anyway.

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Well. Let $\Bbb R$ be a probability space with measure the uniform $[0,1]$ distribution. Let $f:x\mapsto -x$, (which is a random variable that doubles as an injective function) $\Bbb R\to\Bbb R$.

The probability of $(x:f(x)<1)$ is one. But the probability of $(x:x<f^{-1}(1)=-1)$ is zero.

However. If you have a continuous function $I\to\Bbb R$ for some subinterval $I\subset\Bbb R$ which is also injective, then it must be monotonic. So it is either increasing, in which case $f(x)<y\iff x<f^{-1}(y)$ will follow, or it is decreasing, in which case the opposite is true: $f(x)<y\iff x>f^{-1}(y)$.