image of the complement subset of complement of image

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?

Solution 1:

If the statement would be true then $H=f(M)\cup f(M)^c=f(M)\cup f(M^c)$.

This states that $f$ is surjective

Conversely if $f$ is surjective then for every $y\in H$ we can find some $x\in G$ with $y=f(x)$.

If moreover $y\in f(M)^c$ then evidently $x\in M^c$, hence $y=f(x)\in f(M^c)$.

Final conclusion: the statement is true if and only if $f$ is surjective.

Solution 2:

@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