-3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Which definition are you using for continuity? $\endgroup$ May 18 '20 at 10:12
  • $\begingroup$ T is continuous iff the preimage of an open set is an open set $\endgroup$
    – Giulia B.
    May 18 '20 at 10:13
  • $\begingroup$ Then it is pretty straightforward. $\endgroup$ May 18 '20 at 10:14
  • 1
    $\begingroup$ What have you tried? $\endgroup$ May 18 '20 at 10:17
1
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ "Invertible is by definition bijective." It depends on the definition! See math.stackexchange.com/a/2415856/43608 $\endgroup$ May 18 '20 at 11:11
  • $\begingroup$ It's right my proof up? $\endgroup$
    – Giulia B.
    May 18 '20 at 15:41
  • $\begingroup$ T inverse is a map from Y to X @GiuliaB $\endgroup$ May 19 '20 at 6:27
  • $\begingroup$ I've correct the solution using the preimage. Is it right now? $\endgroup$
    – Giulia B.
    May 19 '20 at 7:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.