Injective Bijective Surjective Determine if each one of the following functions $f : A \rightarrow B$ is injective (one–to–one), surjective (onto), bijective, or none of these:
How could these be solved? I'm very confused by the notation at the end in particular. I know what null set and power sets are but the notation at the end with the f() is tricking me. 


*

*$A\neq \emptyset,B=\mathcal{P} (A),f(a)=\{a\}$.

*$A = B = \mathcal{P}({a,b,c,d}), f(X) = X$.

*$A = B = \mathcal{P}({a,b,c,d}), f(X) = X \bigcup \{a,b\}$. 

*$A = B = \mathcal{P}({a,b,c,d}), f(X) = X \bigcap \{a,b\}$.

 A: These define 4 functions, each named $f$, on the pattern $f:A\to B$, i.e. here $A$ denotes the domain and $B$ the codomain (i.e. the set of possible values) of function $f$.


*

*The domain of our first function $f$ is any nonempty set $A$, the codomain is its power set, and $f$ maps an element $a$ to its one-element set $\{a\}$ in the powerset. 
This function is injective and not surjective, hence neither bijective.

*The second function is the identity function ($f(X)=X$) on the 16 element power set $P(\{a,b,c,d\})$. It is bijective.


Can you go on with the other ones?
A: *

*$A\neq \emptyset,B=\mathcal{P} (A),f(a)=\{a\}$.
If $|A|=1$ then $f$ is injective by definition. Assume $|A|>1, \ a\in A,b\in 
 A$ and $a\neq b$. Then $f(a)=\{a\}\neq \{b\}=f(b)$, so $f$ is injective. 
Power set of any set $S$ satisfy following equation:
$$ |\mathcal{P}(S)|=2^{|S|}$$
For example $|\{a,b,c,d\}|=4$ and $|\mathcal{P}(\{a,b,c,d\})|=2^4=16$.
Now $f(A)=\{\{a\}:a\in A\}$ - $f(A)$ is obtained by putting every element 
$a\in A$ to unique set $\{a\}\in \mathcal{P}(A)$. Therefore $|A|=|f(A)|$. 
But $A\neq \emptyset$ so $1\le |A|<2^{|A|}=|\mathcal{P}(A)|=B$. Inequality 
$|f(A)|<|B|$ means, that $f$ is not suriection, neither bijection (not even 
when $|A|=1$!).

*$A = B = \mathcal{P}(\{a,b,c,d\}), f(X) = X$. This function could be redefined as follow:
$$f:\mathcal{P}(\{a,b,c,d\})\ni X \mapsto X \in \mathcal{P}(\{a,b,c,d\})$$
This kind of function is called identity, and it is injective and surjective (thus bijective).
Try to prove that functions (3.) and (4.) are not injective, neither surjective. Observe that in both cases domain $\mathcal{P}(\{a,b,c,d\})$ is the same as codomain.
Sidenote: nullset is subset of any set $S$, so in particular $\emptyset \subset\mathcal{P}(S)$.
