strictly increasing functions and bijection

Question

$f $ is strictly increasing \begin{align} & \rightarrow f^{-1} \text{is strictly increasing}\tag {i} \\ & \rightarrow f^{-1} \text{is strictly decreasing}\tag {ii} \\ & \rightarrow f \quad \text{is injective}\tag {iii}\\ & \rightarrow f \quad \text{is surjective}\tag {iv}\\ & \rightarrow f^{-1} \text{is bijective}\tag {v} \end{align}

which statements are true?

(iii) $f$ is injective since $x<y\rightarrow f(x)<f(y)$

(iv) I believe it doesn't need to be surjective. $f:\Bbb N\to \Bbb N$ with $f(x)=x+1$

(v) since $f$ isn't bijective we cannot talk about $f^{-1} $ function

Answer 1

You almost got everything, the answers are as follows:

• $f$ is strictly increasing $\Rightarrow$ $f^{-1}$ is strictly increasing (is correct)

I would prove this by contradiction. So assume $f^{-1}$ would not be strictly increasing, then you can construct a contradiction pretty straightforward.

• $f$ is strictly increasing $\Rightarrow$ $f^{-1}$ is strictly decreasing (is wrong)

Just take $f(x)=x$ or the use the proof of the first statement

• $f$ is strictly increasing $\Rightarrow$ $f$ is injective (is correct)

Just use the property you already mentioned and the definition of injectivity.

• $f$ is strictly increasing $\Rightarrow$ $f$ is surjective (is wrong)

Your example works or take $f:\mathbb{R}\to\mathbb{R},f(x)=e^{x}$

• $f$ is strictly increasing $\Rightarrow$ $f^{-1}$ is bijective (is wrong)

Generally we have that: $f$ is bijective $\Leftrightarrow$ $f^{-1}$ is bijective.

Because $f$ is not surjective we actually have no inverse function $f^{-1}$ on the whole domain. So all answers above are understood as the restriction of the domain of $f^{-1}$ where it is actually defined, so basically the image $im(f)$ of $f$.

Nevertheless, we now could make $f$ become a bijective function if we would restrict the image space to the image $im(f)$ of $f$. If we do that, then we'd have that $f$ is bijective as also $f^{-1}$.

Answer 2

Assuming no Asymptotic discontinuities if $f(x)$ is Strictly increasing Then $f'(x)>0$.

Letting $$g(x)=f^{-1}(x)$$ we have

$$f(g(x))=x$$ differentiating Both sides

$$f'(g(x))g'(x)=1$$ $\implies$

$$g'(x)=\frac{1}{f'(g(x))}$$ and since $f'$ is Positive

$g'\gt 0$, hence $f^{-1}$ is Increasing.