Menezio

perhaps are you italian?

have a nice day

bye bye

and yes there is no problem with mail

Yep, it is

like also this one

I told you that because i see in your profile that you forget to accept answers

other people will use this chat for speaking about differetial geometry

For the chat no problem, we can leave and there is no problem

I mean, When you ask a question in Stack and people wrote answers, you can accept one of them as the "full answer" to your question. You can do that in your question "submersion or not" page.

stackoverflow.com/help/someone-answers

I mean you can accept my answer

Nope i did not mean that ahahah

thanks ---> you're welcome

you're welcome

byebye, have a nice day

remember to convalidate the aswer

[email protected]

my email is

mmm

exactly

good

so it is invertible and then surjectiv

exactly: the matrix is always a number different from 0

(in our case)

yes, so the matrix is ...

Well for a fixed point $x$ what is the matrix representing the jacobian (i.e. the differential at point x)?

to be a submersion you have to say, doing always this thing, that for all element of $R$ the differential at that point is surjective

so the differential at point $0$ is surjective

Exactly

We are asking for the surjectivity of this matrix

Yes, is the matrix $1\times 1$ with the entry equal to one

well the question now is: what is $df_0$?

Fixed a point $x$ in the domain, now we can say what is the map $df_x$

you can speak about differential when you fix a point.

not just "df" but "df_x"

The map $f:R to R$ is just f = e^x and it is not surjective

They are two different maps.

Tell if I said something strange for you

yes: \neq correspond to =/=

different from

but $x=3$ is just an example; the matrix you have is always invertible (and then surjective) because $e^x \neq 0$ for all $x$

Taking for example $x=3$, the matrix $1\times 1$ with entry $e^3$ is invertible, and it is a surjective map

so

you are asking if this matrix is a surjective linear map for all x in the domain

the Jacobian is the matrix $1 \times 1$ with the entry $e^x$

the differential at a point $x$ coincides, in this exercise, with the jacobian of the map $f$

Ok, well

If you want to think like "calculus" you try to understand that for all $x$ in the domain, the Jacobian valued at the point $x$ is a surjective map