$T: \mathbb{Q}[X]\mapsto\mathbb{Q}[X]$. Proving $T$ is injective and not surjective.

Question

Given $T: \mathbb{Q}[X]\mapsto\mathbb{Q}[X]$, $T(f)=X\cdot X\cdot f - (X+1) \cdot f'$. Prove $T$ is injective and not surjective.

I try to prove this by applying the definition of injectiveness but I cant achieve this. I think I may be missing an important property of Rational Polynomial Functions.

Answer 1

1. Observe that $T$ is a linear map, as $f\mapsto g\cdot f$ and $f\mapsto f'$ are linear, so for injectivity it is enough to check that $T(f)=0$ implies $f=0$.
2. To see, it is not surjective, look at the coefficient of $X$ and the constant term of a general $T(f)$, and deduce that no nonzero constant polynomials are in the image of $T$.