Prove that $f : [a,b] \rightarrow \mathbb{R}$ is a bijection from $[a, b]$ to $[f(a), f(b)]$ I'm a 1st year mathematics student, and in my analysis class I'm having trouble with proving the following:
Let $a < b \in \mathbb{R}$, and let $f : [a,b] \rightarrow \mathbb{R}$ be a continuous and strictly monotone increasing function.
Prove that:
1) $f$ is a bijection from $[a, b]$ to $[f(a), f(b)]$
2) $f^{-1} : [f(a), f(b)] \rightarrow [a, b]$ is continuous and strictly monotone increasing
So far, I think I proved 1):
Injectivity:
Take $x, y \in [a,b]$. If $f(x) = f(y)$, then $x = y$, because $f$ is strictly monotone increasing.  
Surjectivity:
Take $x \in [f(a), f(b)]$. From the definition of this interval directly follows that $x = f(a_1)$ for some $a_1 \in [a, b]$. So f is surjective, and hence injective
Is this prove correct? because it looks a little bit too simple to me.
For 2), I have no idea on how to prove it, except that I think I have to use the epsilon-delta definition of continuity to prove that $f^{-1}$ is continuous.
Could you please tell me iff my prove for 1) is correct and show me how to do 2)?
Thanks in advance!
 A: This is a fairly simple question, and questions like these are often asked by professors at the beginning of one's mathematical path, so that's what I'll assume. By that assumption, I will also presume that the level of strictness in your proof is similar to the typical level.

For injectivity, you will need to be more specific. Did you prove a theorem that a strictly monotone function is injectove? If so, you need to cite that theorem. If not, then you still did not show that $f$ is injective.
It is not enough to say "because it is monotone". You need a sentence or two more to explain what you did.
For (2), your proof is completely false. From the definition of the interval, $x\in[f(a),f(b)]$, it does not follow that $x=f(a_1)$ for some $a_1$. From the definition of the interval, it only follows that $$f(a)\leq x\leq f(b)$$
This is because $[f(a), f(b)]$ is simply defined as
$$[f(a), f(b)] = \{x\in\mathbb R: f(a)\leq x \leq f(b)\}.$$
A: For the surjectivity it's not right.
Let $y\in f([a,b])$. Then there is an $x\in[a,b]$ such that $y=f(x)$. By the fact that $g$ is increasing, $y\in[f(a),f(b)]\subset f([a,b])$. The fact that $[f(a),f(b)])$ is obvious by the intermediate value theorem.
A: For injectivity: Assume $f(x) = f(y)$. Either $x < y$, $x = y$ or $x > y$. If $x < y$ then $f(x) < f(y)$ by monotonicity, which is a contradiction. If $x > y$ then $f(x) > f(y)$ by monotonicity, which is a contradiction. So $x = y$.
For surjectivity: Let $i \in [f(a),f(b)]$. By the intermediate value theorem, there is an $x \in [a,b]$ such that $f(x) = i$.
Monotonicity of $f^{-1}$: Let $u < v$. $u, v \in [f(a), f(b)]$. We know that there are $x$ and $y$ such that $f(x) = u$ and $f(y) = v$. If $x \geq y$ then we would have that $f(x) \geq f(y)$ because of the monotonicity of $f$, which is a contradiction. So $x < y$.
Continuity of $f^{-1}$: Let $y \in [f(a), f(b)]$. Let $\epsilon > 0$. We have that there is an $x$ such that $f(x) = y$. There is an $l$ such that $f(l) = y - \epsilon$ (by surjectivity) and a $u$ such that $f(u) = y + \epsilon$. For any $k$ in $[l,u]$, we have that $f(l) < f(k) < f(u)$ (by monotonicity) so $|f(k) - y| < \epsilon$. In this case our $\delta$ is $\frac{u-l}{2}$.
