Ytukyg

Suppose $f : (a,b) → (c,d)$ is a strictly increasing onto function. Prove that there exists a $g: (a,b) → (c,d)$, which is also strictly increasing and onto, and $g(x) < f(x)$ for all $x ∈ (a,b)$.

I have tried defining $g$ as $g(x) = f(x) - k$ for some small $k$, but that does not seem like the right answer.

You can't use $g(x) = f(x)-k$ because that won't have image $(c,d)$ if $k ne 0$ (because $f(x)$ does have image $(c,d)$).

But you can do something similar. First change coordinates on $(c,d)$, using the coordinate change function
$$h : (c,d) to mathbb R
$$
given by
$$h(x) = tanleft(pileft(frac{x-c}{d-c}right)-frac{pi}{2}right)
$$
This function $h : (c,d) to mathbb R$ is strictly increasing and onto, hence its inverse $h^{-1} : mathbb R to (c,d)$ is also strictly increasing and onto.

Now define
$$g(x) = h^{-1}bigl(h(f(x))-k)bigr)
$$

Hint: this is easy if $a = c = -infty$ and $b = d = +infty$. You can transform the problem for finite $a$, $b$, $c$ and $d$ into the easy case using the functions $tan$ and $arctan$.