How to prove if a function is bijective?

I am having problems being able to formally demonstrate when a function is bijective (and therefore, surjective and injective). Here's an example:

How do I prove that $g(x)$ is bijective?

$$\begin{align} f &: \mathbb R \to\mathbb R \\ g &: \mathbb R \to\mathbb R \\ g(x) &= 2f(x) + 3 \end{align}$$

However, I fear I don't really know how to do such. I realize that the above example implies a composition (which makes things slighty harder?). In any case, I don't understand how to prove such (be it a composition or not).

For injective, I believe I need to prove that different elements of the codomain have different preimages in the domain. Alright, but, well, how?

As for surjective, I think I have to prove that all the elements of the codomain have one, and only one preimage in the domain, right? I don't know how to prove that either!

EDIT

f is a bijection. Sorry I forgot to say that.

Answer

The way to verify something like that is to check the definitions one by one and see if $g(x)$ satisfies the needed properties.

Recall that $F\colon A\to B$ is a bijection if and only if $F$ is:

1. injective: $F(x)=F(y)\implies x=y$, and


2. surjective: for all $b\in B$ there is some $a\in A$ such that $F(a)=b$.

Assuming that $R$ stands for the real numbers, we check.

Is $g$ injective?

Take $x,y\in R$ and assume that $g(x)=g(y)$. Therefore $2f(x)+3=2f(y)+3$. We can cancel out the $3$ and divide by $2$, then we get $f(x)=f(y)$. Since $f$ is a bijection, then it is injective, and we have that $x=y$.

Is $g$ surjective?

Take some $y\in R$, we want to show that $y=g(x)$ that is, $y=2f(x)+3$. Subtract $3$ and divide by $2$, again we have $\frac{y-3}2=f(x)$. As before, if $f$ was surjective then we are about done, simply denote $w=\frac{y-3}2$, since $f$ is surjective there is some $x$ such that $f(x)=w$. Show now that $g(x)=y$ as wanted.

---

Alternatively, you can use theorems. What sort of theorems? The composition of bijections is a bijection. If $f$ is a bijection, show that $h_1(x)=2x$ is a bijection, and show that $h_2(x)=x+2$ is also a bijection. Now we have that $g=h_2\circ h_1\circ f$ and is therefore a bijection.

Of course this is again under the assumption that $f$ is a bijection.