Clarification about the definition of surjectivity

Question

Related to the question Proof that Laplacian is surjective $\mathcal{P}^n\to\mathcal{P}^{n-2}$, I know in general that the surjectivity is defined to be : $\forall f \in \mathcal{P}^{n-2}$, $\exists \hat{f} \in \mathcal{P}^n$ such that $\Delta \hat{f}=f$. However, in the uncookedfalcon's answer, he seems using another definition of surjectivity. Is there anyone could explain to me how he proves the surjectivity of $\Delta : \mathcal{P}^k \to \mathcal{P}^{k-2}$?

Thanks in advance!

Answer 1

They are using the same definition of surjectivity, but also implicitly using an important lemma:

If $f: V\rightarrow W$ is a linear function between two vector spaces, and for some basis $B$ of $W$, we have $\forall b\in B\exists v\in V(f(v)=b)$, then $f$ is surjective.

This is a good exercise. HINT: let $w\in W$; we want to find $v\in V$ such that $f(v)=w$. Since $B$ is a basis for $W$, we can write $w$ as a linear combination of elements of $B$. Do you see how to proceed from here?

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I should note that this isn't really a fact about vector spaces, but general algebraic structures:

If $f: A\rightarrow B$ is a structure-preserving map (homomorphism) between two algberaic structures $A$ and $B$, and $f(A)$ contains some "generating set" of $B$, then $f$ is surjective.

This is of course a bit informal, but it can be made formal (and proved!) in a number of ways.

Answer 2

Ze is using the same definition of surjectivity.

However, ze is also using the fact that $\Delta: \mathcal P^k \to \mathcal P^{k-2}$ is a linear map between vector spaces. Therefore, it's enough to pick a basis for $\mathcal P^{k-2}$ and check that each of the basis elements is in the image of $\Delta$. This is enough to show that $\Delta$ is surjective, because you started by picking a basis!