TITLE: image of the complement subset of complement of image
QUESTION [10 upvotes]: Suppose $f : G \to H$.
Is it true that $f(M)^c \subset f(M^c)$ for $M \subset G$? Is the result ever true—i.e., is it true if we impose the condition that $f$ is bijective?

REPLY [2 votes]: @Dude (maybe 4 years too late but still useful), if you impose that f is surjective and injective then you can conclude that both sets are equal. You can show the missing subset property "$\supseteq$" as follows: 
Let be y $\in f(M^c) \Rightarrow f^{-1}(y) \in M^c$ (this is because f is injective and y has exactly one preimage) $\Rightarrow f^{-1}(y) \notin M \Rightarrow y \notin f(M) \Rightarrow y \in f(M)^c$. The other subset property "$\subseteq$" was already proven by drhab