Let $P=\{p(x)$ | $p(x)$ is a polynomial of degree $n$, $n \in \Bbb Z^+\cup\{0\} $ with coefficients in $\Bbb R \}$. Define $f : P\rightarrow P$ 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?
Best Answer
To see that $f$ is a surjection we take an arbitrary element $y$ in $P$ and show that $\exists x\in P$ 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 $p\in P$, 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.