Inequality and inverse function Is it possible to affirm that if $B\subseteq \mathbb{R}$ 
$$f: \mathbb{R} \rightarrow B $$ and 
$$g: \mathbb{R} \rightarrow B$$ are invertible functions and 
$$f(x) \leq g(x), \forall x \in \mathbb{R} \tag 1$$
then 
$$g^{-1}(x) \leq f^{-1}(x) , \forall x \in B\tag 2$$ 
If so, how can it be proved? If it's false, is there a counterexample?
In this problem, suppose $f(\mathbb{R}) = g(\mathbb{R})$.
If $f$ and $g$ are continous then this is equivalent to both are either strictly increasing or both are strictly decreasing, and the inequalities  $(2)$ will hold.
Intuitively, the proposition seems true to me. I couldn't conceive a counterexample.
 A: As a hint: it is true if $f$ is increasing (in the sense that $x \le y$ iff $f(x) \le f(y)$) and $f(\mathbf{R}) \subseteq g(\mathbf{R})$. For example $f(x) = x$ and $g(x) = e^x$ then $\log x \le x \le e^x$ whenever either inequality makes sense. This isn't hard to prove.
You can use the hypotheses here to construct a counter example to what you've stated.
Something else to think about: saying that $f(x) \le g(x)$ means that $(x, f(x)$ sits below $(x, g(x)$. When you take inverses you reflect in the line $y = x$, this means that $(f(x),x)$ sits to the left of $(g(x),x)$. Why isn't this the same thing as saying that $g^{-1}(x) \le f^{-1}(x)$?

Edit (June 25, 2017):
Since this is true when $f$ is increasing, we take a decreasing $f$ for a counterexample. Let $f(x) = - x$ and let $g(x) = f(x) + 1$. Then $f(x) < g(x)$ for all $x$ but $f^{-1}(x) = f(x) < g(x) = g^{-1}(x)$ so the inequality does not reverse if $f$ is decreasing.

Claim: If $f$ is increasing then the inequality reverses.

Proof: Since $f$ is increasing, $f^{-1}$ is increasing. If $f(x) \le g(x)$ then since $f^{-1}$ is increasing,
$$ f^{-1}(f(x)) \le f^{-1}(g(x)). $$
Since $f^{-1}(f(x)) = x = g^{-1}(g(x))$ we have
$$ g^{-1}(g(x)) \le f^{-1}(g(x)) $$
or $$g^{-1}(y) \le f^{-1}(y) $$ where $y = g(x)$. Since every $y$ is of this form, the inequality holds for all $y$. $\square$

Claim: If $f$ is decreasing then the inequality does not reverse.

Proof: The same except now $f^{-1}$ is decreasing so the inequality reverses at this step:
$$ f^{-1}(f(x)) \ge f^{-1}(g(x)). $$
So we have
$$ g^{-1}(y) = f^{-1}(f(x)) \ge f^{-1}(y) $$
where $y = g(x)$. $\square$
