Jens Renders

what is your definition then

otherwise dx/dy has no meaning for a function y = f(x)

so f^-1(y) has to be differentiable

and then we take derivativ to y

x = f^-1(y)

to have x as a function of why we need f^-1 to solve for x

so if we start from y = f(x)

the notation dx/dy means, write x as a function of y and then take derivative to y

and dx/dy has also no meaning then

indeed

so we can only write dx/dy if x is a function of y

then we can write dy/dx

and if y = f(x)

or df/dx

for a function f(x), its derivative to x is denoted df(x)/dx

if f is a function of x

dx/df is wrong notation

does (f^-1)' exist?

if f is a function such that there is no inverse f^-1

in other notation you are asking

you can not take a derivative of a function that does not exist

that is a nonsensical question

You are asking, if f has no inverse, can its inverse have a derivative?

@VivaanDaga Can what derivative still exist?

@VivaanDaga but $\frac{dx}{dy}$ is by definition the derivative of $x = f^{-1}(y)$ to $y$.

@VivaanDaga What do you mean an example? If $f^{-1}$ is not differentiable then the theorem statement does not make sense. It contains the derivative of $f^{-1}$: $\frac{1}{dx/dy} = 1/(f^{-1})'(y)$ by definition.

@amWhy Okay thanks!

I do want you're help but not for money, I help people on here all the time (that's what this site is for right)

how am I paying your family

what methods don't work for what?

Okay took me long enough to realise that you are not making any sens at all. I just tough you would be sincere because of your rep but appareantly that doesn't say it all

I am a belgian math student, nothing to hide

Stop what experiments? I'm doing my best to provide the information needed for this question to be clear, but you seem to talk nonsense to me

@mathreadler what information do you need? I've linked the book and provided the page number, you can have a look there. If you want me to state some definitions in the question, just tell me which ones. I'm not sure which concepts and notations are common knowledge in this area

could you please cleary explain what you are trying to say?

you are acting quite strange, especially for someone with 9k reputation

@mathreadler Nothing -more- to contribute? then what have you contributed? I don't get the point of your comments

@mathreadler I don't get what you are trying to imply? I know that this is a good book, I know that the theorem is false for arbitrary $n$ and $m$ (that is obvious), so i've been going trough the argument again and again and not finding the point where this is used. Hence my question

@mathreadler I've gone trough the whole argument. every step seems to work for arbitrary $n$ and $m$. This is ofcourse impossible, hence my question

examples of functors along with proofs of the fact that they are functors, the internet is full if it

maybe look up some examples of it

by logic

by carefully defining what a functor is. That is what is done in the link a cited, and if you want to check wether a certain map is a functor, you check the definition, it's that simple

And what other codomains did you have in mind? Sets as objects and diffrent morfisms? Then those morfisms arent funtions so it cannot violate the definition of functions obvioulsy

I litterally cited the definition, please read it. $F(.)$ does not associate $F(A)$ and $F(B)$, it associates a morfism $f:A\rightarrow B$ to another morfism $F(f): F(A) \rightarrow F(B)$. You try to define F(f) = f, but f is a morfism from A to B not from -A to -B.

The dual (opposite) category of set is a valid category, the morfisms just arent functions, thats not a requirement to be a category

@MPitts yes, $F$ is not a function, since it's domain and codomain aren't sets. However, we are talking about $F(f)$ wich is according to the definition i cited a morfism. The morfisms in Set are functions, so yes, the definition does say that $F(f)$ should be a function and yes, it does say that it should be a function between $F(A)$ and $F(B)$.

@MPitts And your example in the dual category doesn't really make any sense. In the dual category of Set, the morfisms are no longer functions

@MPitts Have you read it? I quote: "associates to each morphism $f:X \rightarrow Y$ in $C$ a morphism $F(f): F(X) \rightarrow F(Y) $ in $D$ such that..."

@MPitts It certainly is a proof that $F$ is not a functor, because the demand that $F(f)$ should be a map from $F(A)$ to $F(B)$ is part of the definition of a functor. $F$ doesn't satisfy this, so by definition is not a functor. I've editted in a link to this definition