Determining if $Z$ is injective or surjective. Help starting a proof I have $\mathfrak P(\mathbb{R})$ being the set of all subsets of $\mathbb{R}$, meaning $\mathfrak P(\mathbb{R}) = \{X|X\subseteq \mathbb{R}\}$. I then have $F$ being the set of all functions $\mathbb{R} \rightarrow \mathbb{R}$. Define $Z:F \rightarrow \mathfrak P(\mathbb{R}) $ by $Z(f) = \{x \in \mathbb{R}|f(x) = 0 \}$ and I am trying to work out if Z is injective or surjective.
I feel as though mapping all functions, onto an infinite number of infinite sized subsets would surely be injective and not surjective. How can I start a proof of this?${}$
I still don't get it. 
 A: You need to think about what the map $Z$ means. Saying that $Z$ is surjective means that for any subset of $\mathbb{R}$, there is some map $\mathbb{R}\to\mathbb{R}$ that has that subset as its set of zeros. Saying that $Z$ is injective says that if two functions from $\mathbb{R}$ to $\mathbb{R}$ have the same set of zeros, then they are equal. If you understand those statements, you should easily be able to answer the question.
A: I'll furnish the details to rogerl's answer.
Surjectivity: Note that $Z$ can either be surjective or not surjective. To show that it is surjective one needs to show that for ANY subset $X\subseteq \mathbb{R}$, we have a function, $f$ say, such that $f(x)=0$ for all $x\in X$.
The function $f(x)=\begin{cases}
0&\text{if } x\in X\\
1&\text{if } x\notin X\\
\end{cases}$ springs to mind.
Injectivity: If $Z$ is injective, then if $f,g:\mathbb{R}\rightarrow\mathbb{R}$ have the same set of zeros, then it must be that $f=g$. Consider $f(x)=x$, $g(x)=x^2$, they have the same set of zero, namely $0$ itself, but they are not the same function, hence, $Z$ is not injective.
