Short question, what is a set mapping? I find it difficult using a set as an argument. I am struggling with measure theory at the moment and the example is a function
$f: S \to T$ , s.t.
$f^{-1}(B) = \{s \in S; f(s) \in B \}$ with $B \subset T$
If someone could explain this.
Thanks.
$\endgroup$3 Answers
$\begingroup$Let it be that $S$ and $T$ are sets and that $f:S\to T$ is a function.
We have the powersets $\mathcal{P}(S)$ and $\mathcal{P}(T)$.
Function $f$ induces/determines a function $\mathcal{P}(S)\to\mathcal{P}(T)$ that is prescribed by: $$A\mapsto\{f(x)\mid x\in A\}$$ The output is abbreviated by $f(A)$ so we have: $$A\mapsto f(A)$$
Function $f$ induces/determines a function $\mathcal{P}(T)\to\mathcal{P}(S)$ that is prescribed by: $$B\mapsto\{x\in S\mid f(x)\in B\}$$ The output is abbreviated by $f^{-1}(B)$ so we have: $$B\mapsto f^{-1}(B)$$
Caution: we are not dealing with any inverse map of $f$ as the notation seems to suggest.
$\endgroup$ $\begingroup$Instead of looking at the inverse finction it is easier to look at the function $f$ and understand set mappings there first. The function of a set $B$ is then the set of all elements hit when elements in $B$ are mapped using $f$, or in other words it is the range of the function restricted to the specific set.
For instance, assume $x,y\in S$, then $f(\{x,y\}) = \{f(x),f(y)\}$. Especially notice that if $f(x)=f(y)=a$ (for some element $a$), then we may write $f(\{x,y\})= a$.
Now if we want to study the inverse function, we instead need to look at what elements are mapped to a certain set. Let $f(x)= x^2$. Then $f^{-1}(4)= \{2,-2\}$ while $f^{-1}([0,4]) = [-2,2] $, since all elements in the intervall $[-2,2]$ are mapped in the interval $[0,4]$, and these are the only elements which are mapped there.
$\endgroup$ 1 $\begingroup$A set mapping is a map between sets. This carries remarkably little information on its own.
A function $f: S \longrightarrow T$ is a way of associating to each $s$ in $S$ some $t$ in $T$. The $t$ associated to an $s$ is usually denoted by $f(s)$.
Alternately, a function $f: S \longrightarrow T$ is a particular relation, i.e. a set of ordered pairs $(s, t)$ with $s$ in $S$ and $t$ in $T$, with the property that there is exactly one ordered pair containing $s$ for each $s$ in $S$.
In your question, you've mentioned $f^{-1} B$, which might be read as "the preimage of $B$". You have defined this as those elements in $S$ that are sent to elements in $B$ by the function $f$ (or rather, elements in $S$ associated to elements in $B$ under the relation given by $f$).
For instance, suppose $g: \mathbb{Z} \longrightarrow \mathbb{Z}$ is the map $$ g(n) = \begin{cases} 0 & n \text{ is even} \\ 1 & n \text{ is odd} \end{cases}$$
Then you might ask what is $g^{-1} 0$? And it is exactly the even integers.
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