- #1
jeff1evesque
- 312
- 0
Hello,
I just cracked open this abstract algebra book, and saw a problem I have no idea how to solve.
Instruction:
Determine whether the given map [tex]\phi[/tex] is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? (Note: F is the set of all functions f mapping [tex]R[/tex] into [tex]R[/tex] that have derivatives of all orders.Problem:
[tex]<F, +> with <F, +>[/tex] where [tex]\phi(f)(x) = \int_{0}^{x}f(t)dt[/tex]
Definition:
An isomorphism is bijective mapping (one to one, and onto), and also homomorphic.
Solution:
The answer is no because F does not map F onto F? I am very bad at this, what does the following show,
[tex]\phi(f)(0) = 0[/tex]
Does that show it is not onto? Can someone explain to me why this is not an isomorphism (I am terrible at this- all I know is the definition from wikipedia of bijective mapping and homomorphis- but I cannot apply it to these problems)?
I just cracked open this abstract algebra book, and saw a problem I have no idea how to solve.
Instruction:
Determine whether the given map [tex]\phi[/tex] is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? (Note: F is the set of all functions f mapping [tex]R[/tex] into [tex]R[/tex] that have derivatives of all orders.Problem:
[tex]<F, +> with <F, +>[/tex] where [tex]\phi(f)(x) = \int_{0}^{x}f(t)dt[/tex]
Definition:
An isomorphism is bijective mapping (one to one, and onto), and also homomorphic.
Solution:
The answer is no because F does not map F onto F? I am very bad at this, what does the following show,
[tex]\phi(f)(0) = 0[/tex]
Does that show it is not onto? Can someone explain to me why this is not an isomorphism (I am terrible at this- all I know is the definition from wikipedia of bijective mapping and homomorphis- but I cannot apply it to these problems)?