I did this exercise. I am not sure if my surjective proof is right – is it good? In fact, I'm not even sure what exactly am I trying to prove (I know that surjective means that each element in the codomain should have a preimage in the domain, but I don't really see how to prove that – I was just mimicking some example I had seen before). Could you give more insight on how to prove a function surjective?
The rest of the exercise is here as well, in case I did a related mistake.
Let $f:[0,+\infty[ \rightarrow [1,+\infty[$ defined by
$$f(x) = x^2 + 1$$
Demonstrate that it is bijective.
Injective
$$f(a) = f(b)$$
$$a^2+1 = b^2 + 1$$
$$a^2 = b^2$$
Since $a,b >= 0$, it is safe to affirm that
$$a = b$$
Surjective
$$b = f(a)$$
$$b = a^2 + 1$$
$$b – 1 = a^2$$
$$a = \sqrt{b – 1}$$
Calculate $f^{-1}$ and verify that $f \circ f^{-1}(x) = x$.
$$f^{-1}(x) = \sqrt{x – 1}$$
And
$$f \circ f^{-1}(x)$$
$$f(f^{-1}(x))$$
$$(\sqrt{x – 1})^2+1$$
$$x – 1 + 1$$
$$x$$
Best Answer
Yes, you're exactly correct, for we have
$$f(a) = a^2 + 1 = \Big(\sqrt{b - 1}\Big)^2 + 1 = b - 1 + 1 = b$$
as desired. This is the important thing to prove.
The general strategy to show that a function is surjective is two-fold:
Find a candidate for the preimage by working backwards (i.e. scratchwork)
Show that the candidate actually works
Now in a case such as this where all the steps in solving for $a$ in terms of $b$ are reversible, these steps are really equivalent.