If a function maps $A$ to its PowerSet, is it Surjective? Given an arbitrary set $A$, let $F : A \rightarrow 2^A$ be the function defined for
all $a \in A$ by
$f(a) = \{a\}$
If $A$ maps to its power set, does this make $F$ surjective?
If somebody could help to prove this that would be very helpful
 A: No it's not surjective because $ \emptyset $ is not in the range of $f$
A: There is no surjective map from $A \to 2^A$ as Cantor has taught us with his diagonal argument.
A: No. There is no $x$ for which $f(x)=\emptyset$. Surjective means it hits everything in $2^A$ which it doesn't because $\emptyset \in 2^A$.
$f$ is injective though.
A: Let's assume $A$ itself is non-empty so the proposed mapping $f$ has non-empty image. Elements of the power set of $A$ are subsets of $A$. The power set therefore comprises the empty set, "singleton" sets containing exactly one element, sets containing two elements, or three, etc.; if $A$ itself contains infinitely many elements, then $A$ also has subsets containing infinitely many elements, and each is an element of the power set.
In order for a function with codomain $2^{A}$ to be surjective, every subset of $A$ must be in the image.
The image of the mapping $f(a) = \{a\}$ consists of the singleton subsets of $A$. Not every subset of $A$ is a singleton, so $f$ is not surjective.
Perhaps you're conflating the image of $f$ with the union
$$
\bigcup_{a \in A} f(a) = \bigcup_{a \in A} \{a\} = A?
$$
