Showing this map is injective

Question

Define a map $$\phi : \mathbb{Q}[x] \rightarrow \mathbb{Q}[x]$$ by $\phi(g(x)) = g(x-2)$

It is straightforward to show that this is a homomorphism, but I'm having trouble showing it's surjective & injective.

For surjective, I just said that given $g(x) \in \mathbb{Q}[x]$, then $g(x) = \phi (g(x+2))$, is this enough to show it's surjective?

For injection, I let $\phi (g(x)) = \phi (h(x))$, and aim to show $h = g$. So I get that $g(x-2) = h(x-2)$, does this imply that $h = g$? I'm not sure?

Answer 1

Injectivety:

If $\phi (g(x))= \phi (f(x))$ then $g(x-2)=f(x-2)$ for all $x$ so if we put $x=t+2$ we get $g(t) = f(t)$ so $g(x) = f(x)$.

Surjectivety:

Take any $h(x)$ in $\mathbb{Q}[x]$. So if we take $g(x) := h(x+2)\in\mathbb{Q}[x]$ then $$\phi(g(x)) = g(x-2)= h(x)$$

Answer 2

The ring homomorphism \begin{align} &\psi:\Bbb Q[x]\to\Bbb Q[x]& &f (x)\mapsto f (x+2) \end{align} is the inverse of $\phi $, hence $\phi $ is bijective.