Let $f$ be a function from $X$ to $\mathbb R$ and $g$ be a function from $X$ to $X$. Furthermore, define $h(x) = \sup_{y:x=g(y)}f(y)$.
Then it apparently holds that $$\sup_{x\in X}h(x) = \sup_{x\in X}\sup_{y:x=g(y)}f(y) = \sup_{x\in X}f(x)$$
This is used in the prove of the invariance property of MLEs (c.f. https://stats.stackexchange.com/questions/333737/maximum-likelihood-is-not-re-parametrization-invariant-so-how-can-one-justify-u). I don't understand why the second equality follows. My professor said it is "obvious", however, I don't see why.
Kavi Rama Murthy's objection can be gotten around either by changing the statement to $$\sup_{x \in g(X)} h(x) = \sup_{x \in X} f(x)$$ or defining $h : X \to \overline {\Bbb R}$, with $\sup_{x \in \emptyset} f(x) := -\infty$.
In either case, the statement is true.
Now let $M = \sup_{x\in X} f(x)$. For any $m < M$, there is a $y_m\in X$ with $f(y_m) > m$. Let $x = g(y_m)$. Then $h(x) \ge f(y_m) > m$. Therefore $\sup_{x \in X} h(x) > m$ for all $m < M$. Since $h(x) \le M$ for all $x$, we must have $$\sup_{x \in X} h(x) = M$$