Question

is a strictly increasing onto function a bijection?​

Answer 1

Answer:

bijection to vo Hoti h Jo both one one and onto ho

Answer 2

The answers are as follows:

ff is strictly increasing ⇒⇒ f−1f−1 is strictly increasing (is correct)

I would prove this by contradiction. So assume f−1f−1 would not be strictly increasing, then you can construct a contradiction pretty straightforward.

ff is strictly increasing ⇒⇒ f−1f−1 is strictly decreasing (is wrong)

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

ff is strictly increasing ⇒⇒ ff is injective (is correct)

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

ff is strictly increasing ⇒⇒ ff is surjective (is wrong)

Your example works or take f:R→R,f(x)=exf:R→R,f(x)=ex

ff is strictly increasing ⇒⇒ f−1f−1 is bijective (is wrong)

Generally we have that: ff is bijective ⇔⇔ f−1f−1 is bijective.

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

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