1
$\begingroup$

Suppose I have $B=F^{-1}(A)$, shouldn’t $F^{-1}$ defined on all of $A$ for the inverse map to make sense? Or can we define $B=({{ u\in B|F(u)\in A }})$? My question is regarding a proof in Pressley’s Elementary Differential Geometry. How is $\tilde{\sigma}^{-1}$ defined on all of $\sigma(W)$?enter image description here can someone explain this?

$\endgroup$
  • $\begingroup$ I'm at a loss here what your question is about. Nowhere in the text in your image is there an inverse map being applied to a set at all. And in every case, the map being inverted is bijective, and therefore its inverse is defined on the entire codomain. What exactly are you asking about? $\endgroup$ – Paul Sinclair Jan 15 at 15:03
  • $\begingroup$ My exact question is how is $\tilde{\sigma}^{-1}$ defined on all of $\sigma(W)$? Since $\tilde\sigma(u_0,v_0) = \sigma(u_0,v_0) = P$ it is possible that $\tilde{\sigma}^{-1}$ is defined on some part of $\sigma(W)$ but how can we be sure it is defined on all of it? How is $\sigma(W)$ is the domain of $\tilde{\sigma}^{-1}$? $\endgroup$ – danny Jan 15 at 16:34
  • $\begingroup$ Okay - I missed that instance of an inverse image of a set. Given a map $f$ with domain $D$ and any set $A$, the definition of $f^{-1}(A) = \{ x \in D \mid f(x) \in A\}$. This definition does not put any requirements on $A$. It doesn't matter if $A$ contains points not in the codomain of $f$, much less the image. Those points will not result in anything being added to the inverse image, but this is not a problem for the definition. Note that they are proving that the transition map is smooth where it is defined. $\endgroup$ – Paul Sinclair Jan 16 at 3:08

Your Answer

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

Browse other questions tagged or ask your own question.