if $f$ is injective, then is it right $f(B^c) =f(B)^c$?

Question

Let $f: X \rightarrow Y$, and $A$ is subset of $X$.

Then is it right $f(A^c) = f(A)^c$ if $f$ is injective?

What about the condition "injective" is not involved?

Answer 1

First, $B$ should be a subset of X, not of Y. Yes, injectivity is required but also surjectivity is required for the equality to hold. With injectivity we only have, $ f(B^c) \subset (f(B))^c$ A counterexample would be

$f(x) = tan^{-1}x$

Since taking $ B = \mathbb{R}\setminus{0} $

we have, $f(B^c) = f(0) = 0$

But, $ (f(B))^c = [π/2, \infty) \cup (-\infty, -π/2] \cup {0}$

Answer 2

Why is injectivity needed? Consider $f: \{1, 2\} \to \{0\}$, with $f(1) = f(2) = 0$, and $A = \{1\}$. $f(A) = \{0\}$ and $f(A^c) = \{0\}$ as well.

Why is injectivity sufficient? Say $x \in f(A^c)$, then there is $w \in A^c$ such that $f(w) = x$. But $f$ is injective. So no other thing $w' \in X$ will have $f(w') = x$. In particular, for any $w' \in A$, $w' \not= w$, and thus $f(w') \not= x$. This means $x \not\in f(A)$ and thus $x \in f(A)^c$. Since $x$ is chosen arbitrary, this shows that $f(A^c) \subseteq f(A)^c$.

The other direction is similar.