# T is an open map if and only if $T^{-1}$ is continuous

Let X and Y topological space and $$T:X\rightarrow Y$$ invertible. Prove that $$T^{-1}$$ is continuous if and only if T is an open map.

Is T bijective?

$$\Leftarrow)$$ if $$T$$ is an open map then $$\forall A \subseteq X$$, $$A$$ open set then $$T(A) \subseteq Y$$ is an open set. $$T$$ is an invertible map so $$(T^{-1}(T(A))=A$$ i.e. the preimage of an open set of $$Y$$ is an open set on X and this is the definition of continuity for $$T^{-1}$$?

$$\Rightarrow)$$: $$T^{-1}$$ is continuous so the preimage of an open set in Y is an open set in X. T is invertible so T sends open set in X in open set in Y i.e. T is an open map

• Which definition are you using for continuity? May 18 '20 at 10:12
• T is continuous iff the preimage of an open set is an open set May 18 '20 at 10:13
• Then it is pretty straightforward. May 18 '20 at 10:14
• What have you tried? May 18 '20 at 10:17

If T is open that means it takes open sets in X to open sets in Y. To prove $$T^{-1}$$ is continuous then clearly preimage of an open set in X is open in Y as $$(T^{-1})^{-1}$$ is nothing but T itself which is open by the hypothesis. Similarly the converse follows.