Is $x < y \implies f(x) < f(y)$ equivalent to $x < y \iff f(x) < f(y)$? Given a function $f: \mathbb{R} \to \mathbb{R}$, is $x < y \implies f(x) < f(y)$ equivalent to $x < y \iff f(x) < f(y)$.
If $f(x) < f(y)$, then the contradiction occurs only$_\text{(or is it not the "only" case)}$ when $x > y$. If that happens, then $f(x) > f(y)$ and so the two statements are always equivalent.
Am I missing something? Is it always the case or it changes when we change the injectivity, surjectivity, domain or anything else?
 A: Yes. Suppose that LHS holds. Let $x,y\in\mathbb{R}$. Suppose that $f(x)<f(y)$. We go to prove that $x<y$ by contradiction. Suppose the contrary that $x \not <y$, then $x\geq y$, i.e., $x=y$ or $x>y$.
If $x=y$, we have $f(x)=f(y)$, contradicting to the given condition.
If $x>y$, by the hypothesis (LHS), we have $f(x)> f(y)$, which contradicts to the given condition $f(x)<f(y)$.
Therefore $f(x)<f(y) \Rightarrow x<y$.
A: Suppose $f(a)\gt f(b)$ then we have three possible cases
$a=b$ - well then, by the fact that $f(x)$ is a function we have $f(a)=f(b)$, which is not the presenting case
$a\lt b$ in which case the conditions would give us the contradiction $f(a)\lt f(b)$
Or $a\gt b$, which is the only case which works.
So the reverse implication holds, as required.
Don't be afraid of considering cases when they are not so many in number.
A: Yes, they are equivalent. The key here is that "$<$" is a total order on $\mathbb{R}$.
Assume $f$ satisfies the condition $$(*)\quad \mbox{For all $a,b$}, 
a<b\implies f(a)<f(b)$$ and suppose $f(x)<f(y)$. Clearly this means $\color{red}{\mbox{$x\not=y$, so either $x<y$ or $y<x$}}$. We can't have $y<x$ since then $(*)$ would imply $f(y)<f(x)$, so we must have $x<y$.
This applies to any total order. On the other hand, it breaks down for posets since the implication $x\not=y\implies (x\triangleleft y\mbox{ or }y\triangleleft x)$ no longer needs to hold if $\triangleleft$ is merely assumed to be a partial ordering.
A: If we know that for all $x$ and $y$, $\; x<y\implies f(x)<f(y)$, then we can prove $f(x)<f(y)\implies x<y$ by contraposition. The contrapositive of $f(x)<f(y)\implies x<y$ is
$$
x\ge y\implies f(x)\ge f(y) \, . \tag{*}\label{*}
$$
If $x=y$, then $\eqref{*}$ clearly holds. If $y<x$, then by assumption $f(y)<f(x)$, and so $f(x)\ge f(y)\blacksquare$
