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Let $P={p(x)$ | $p(x)$ is a polynomial of degree $n$, $n in Bbb Z^+cup{0} $ with coefficients in $Bbb R }$. Define $f : Prightarrow P$ where $f(p(x)) =p'(x)$, the derivative of $p(x)$.

Determine whether $f$ is a function, an injection, a surjection, a bijection.

Now I have the solutions, and I understand that it is a function because each polynomial has a unique derivative. And it is not an injection as the antiderivative of a given polynomial is not unique.

However, I do not understand the book's solution for determining whether it is a surjection, nor am I able to come up with one myself. In all honesty, I think I am having trouble understanding the mapping from P to P. The solution states that it is a surjection. Why is this so?

To see that $f$ is a surjection we take an arbitrary element $y$ in $P$ and show that $exists xin P$ such that $f(x)=y$. This is what it means to be surjective; we cover the entire space with the image of $f$ on $P$.

Let $pin P$, i.e. $p$ is some polynomial $p(x)$. Then $p(x)$ has an antiderivative, $q(x) = int p(x)dx$. This antiderivative is a polynomial (easy to check), so it is in $P$. Then $f(q) = p$, and since $p$ was arbitrary, $f$ is therefore surjective.

Here is a more concrete analogy to help you understand what a surjection is.

Imagine you have two lists of names, one for adults and one for children. Imagine that someone says they have a way to assign names from the adult list to names in children list. Their way of assigning the names is surjective if every name is the children list is assigned at least one name in the adult list. This would mean that if you pick a name from the children list as random, there is at least one adult name corresponding to it.

For any polynomial $p (x) $ there exists its integral $P (x) $, for which $p $ is a derivative: $$p (x)=P'(x)$$ and which itself is a real-coefficients polynomial, too $$P(x)in P$$ Hence each $p(x) in P$ is in the image of your function $f$. Then $f$ is surjective.

Think about what the inverse mapping for $f$ would be if indeed it would be a surjection: it would take the derivatives to the original function. Phrased another way, it would take a function to its antiderivative, right?

Well, we know that all polynomials have an antiderivative (which is itself a polynomial), and since the codomain of $f$ is basically the set of real polynomials, we have a pre-image for each polynomial in $P$ - its antiderivative. Thus, $f$ is surjective.