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

Question

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

T is continuous iff the preimage of an open set is an open set — Giulia B., May 18 '20 at 10:13

score 1 · Accepted Answer · answered May 18 '20 at 10:20

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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.

Invertible is by definition bijective.

answered May 18 '20 at 10:20

Devendra Singh Rana

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"Invertible is by definition bijective." It depends on the definition! See https://math.stackexchange.com/a/2415856/43608 – Jean-Claude Arbaut May 18 '20 at 11:11
It's right my proof up? – Giulia B. May 18 '20 at 15:41
T inverse is a map from Y to X @GiuliaB – Devendra Singh Rana May 19 '20 at 06:27
I've correct the solution using the preimage. Is it right now? – Giulia B. May 19 '20 at 07:38

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

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