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 December 19th, 2017, 09:51 PM #11 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 I want to define a "isomorphism of function". And I need to understand the structure of functions that are groups.
December 20th, 2017, 05:22 PM   #12
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Quote:
 Originally Posted by policer I want to define a "isomorphism of function".
As has been mentioned, isomorphisms are generally between groups or other algebraic structures. There aren't isomorphisms of functions. Can you clarify what you are asking?

Quote:
 Originally Posted by policer And I need to understand the structure of functions that are groups.
There are no individual functions that are groups. There are many sets of functions that are groups. Again, can you clarify your intention?

 December 21st, 2017, 02:42 AM #13 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 I think isomorphism of functions can define an operation on function(s) that are(/is) can be drawn the function differently when I use this operation. And for me I is very very interesting to draw a function on this operation.
 December 21st, 2017, 03:07 AM #14 Senior Member   Joined: Sep 2016 From: USA Posts: 357 Thanks: 196 Math Focus: Dynamical systems, analytic function theory, numerics Why does this particular site get so many lunatics?
December 31st, 2017, 07:46 AM   #15
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Quote:
 Originally Posted by Maschke As has been mentioned, isomorphisms are generally between groups or other algebraic structures. There aren't isomorphisms of functions. Can you clarify what you are asking?
Well, it's not quite correct I'm afraid. An isomorphism is in general a certain morphism in a category. Thus if we want an isomorphism between two functions, you'd need to specify a suitable category.

One very interesting category is the comma category. I'm gonna skip all the details, but it suffices to say that an isomorphism between two functions $f:A\rightarrow B$ and $g:C\rightarrow D$ consists of a pair of bijections $(\varphi: A\rightarrow C,\psi: B\rightarrow D)$, where $g\circ \varphi = \psi \circ f$.

Now I'm pretty sure the OP didn't have this in mind and is just a confused student. But perhaps some other people would find this interesting.

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