Implication about homeomorphism

Question

Let $f:(X,\tau_1) \to (Y,\tau_2) $ be a bijection and $A \subseteq X.$ Then TFAE

a)f is a homeomorphism

b) $f(\overline A) = \overline {f(A)}$

c) $f(int(A)) = int(f(A))$

Which means we should show a $\Rightarrow$ b $\Rightarrow$ c $\Rightarrow$ a

How can I show $$f(\overline A) = \overline {f(A)} \Rightarrow f(int(A)) = int(f(A))$$

($\overline A $ is closure of $A$ and $int(A)$ is interior of $A$)

We can say $f$ is closed and continuous from assumption. Namely, $f$ is open and continuous. Thus $f(int(A)) \subseteq int(f(A))$. I have problem about showing $int(f(A)) \subseteq f(int(A))$

Answer 1

Prove that a) and b) are equivalent and then prove that a) and c) are equivalent. a) implies b) is clear. Suppose b) holds. If A is a closed set than b) shows that f(A) is closed too. If U is open then its complement A is closed, so f(A) is closed. But f(A) is the complement of f(U) (because f is a bijection). Hence f(U) is open. The fact that f maps open sets to open sets and closed sets to closed sets shows that a) holds. A similar argument shows that c) is equivalent to a).