$f(S)^c \subseteq f(S^c)$, $\forall S\subseteq A$, if and only if $f\colon A\to B$ is surjective Let $f:A \rightarrow B$. 
The complements indicated below are taken within $A$ or $B$.
I need to prove that
$f(S)^c \subseteq f(S^c)$, $\forall S \subseteq A$, if and only if $f$ is surjective. 
So I need to prove $f(A) = B$ right?
How can I prove this?
 A: You want to prove two things:


*

*If the complement of the image is always contained in the image of the complement, then $f(A)=B$; and

*If $f(A)=B$, then for every subset $S$ of $A$ you have that the complement of the image is contained in the image of the complement.
So, "I have to prove that $f(A)=B$" is only true for half of the problem at issue.
As to how you prove 1 above, well, since the containment holds for all subsets $S$ of $A$, perhaps you can pick a particular subset $S$ of $A$ in some clever way that will tell you that $f(A)$ must be equal to $B$. (HINT: $f(A)=B$ if and only if $f(A)^c$ is... )
For 2, you have to assume that $f(A)=B$ (that $f$ is surjective). Then you need to take an arbitrary subset $S$ of $A$. Then to test whether $f(S)^c\subseteq f(S^c)$ holds, you need to take an arbitrary $b\in f(S)^c$, and show that it must lie in $f(S^c)$; that is, that there exists some $a\notin S$ such that $b=f(a)$. Now go to it.
A: HINT $\rm\ (\Rightarrow)\ $ Put $\rm S\ =\ \ldots\ \ (\Leftarrow)\ $ Consider $\rm\  {\overline {f(S)}} \cap (f(S) \cup f(\overline S))$
A: You need to prove more than that $f(A) = B$. You need to show that if $f$ is surjective, then $(f(S))^c \subset f(S^c)$ for every possible subset $S$ of $A$, and that if $f$ is not surjective, then there is some $S$ for which $(f(S))^c$ is not a subset of  $f(S^c)$.
Largish hint for the first part: If $f$ is surjective then $f(S) \cup f(S^c)$ is all of $B$, regardless of what $S$ is. Use this to prove the contrapositive of the first part.
Hint for the second part: Pick a special subset $S$ of $A$ and show it works for that $S$. 
A: (I assume it is OK to post a full solution to what was long ago probably a homework question.)
Let me provide a complete (but perhaps overly long-winded) proof which does not use separate $\Rightarrow$ and $\Leftarrow$ parts, but only equivalences.
The most complex part seems to be $f[S]^c \subseteq f[S^c]$, so let's try to simplify that: for any $S \subseteq A$,
$$
\begin{align}
& f[S]^c \subseteq f[S^c] \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& \langle \forall y :: y \in f[S]^c \Rightarrow y \in f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $^c$, which means $B$-complement here"} \\
& \langle \forall y :: y \in B \land y \not\in f[S] \Rightarrow y \in f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"logic: rearrange to bring both occurrences of $f[\cdot]$ together"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[S] \lor y \in f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\cup$ -- since we know distribution properties of $\cdot[\cdot]$"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[S] \cup f[S^c] \rangle \\
(*) \; \equiv & \;\;\;\;\;\text{"$f[\cdot]$ distributes over $\cup$"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[S \cup S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"set theory: basic property of $^c$, which here means $A$-complement"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[A] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& B \subseteq f[A] \\
\equiv & \;\;\;\;\;\text{"using $f[S] \subseteq B$ for any $S$, since the range of $f$ is $B$"} \\
& B = f[A] \\
\equiv & \;\;\;\;\;\text{"one of the definitions of surjectivity, using $f : A \to B$"} \\
& f \textrm{ is surjective} \\
\end{align}
$$
Now formally wrapping up, we have
$$
\begin{align}
& \langle \forall S :: f[S]^c \subseteq f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"by the above calculation"} \\
& \langle \forall S :: f \textrm{ is surjective} \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify: $S$ does not occur inside $\forall S$"} \\
& f \textrm{ is surjective} \\
\end{align}
$$
which proves the statement in question.
The key step was $(*)$, and this is the most 'creative' part in an otherwise fairly mechanical proof, provided one is familiar with logic and the definitions and basic properties of set theory and functions.
