Determine whether the given map is an isomorphism

[jeff1evesque]

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 \phi 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 R into R that have derivatives of all orders.Problem:
<F, +> with <F, +> where \phi(f)(x) = \int_{0}^{x}f(t)dt

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,
\phi(f)(0) = 0

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)?

 

[VeeEight]

Can you find a counterexample to show that phi is not one to one? That is, find two distinct functions in F, call them f and g, such that phi(f) = phi(g)

Have you tried checking if it is a homomorphism?

 

[Hurkyl]

what does the following show,
\phi(f)(0) = 0

It shows that every function in the image of phi has the value 0 at 0.

 

[jeff1evesque]

I can't seem to find a counterexample. I mean if I let f(t) = t^{2} and let g(t) equal to anything else, it would seem f(t) = t^{2} \neq g(t), unless t = 0- but that doesn't make the function equal, it just shows for a particular domain value, the functions are equal.

 

[rasmhop]

I can't seem to find a counterexample. I mean if I let f(t) = t^{2} and let g(t) equal to anything else, it would seem f(t) = t^{2} \neq g(t), unless t = 0- but that doesn't make the function equal, it just shows for a particular domain value, the functions are equal.

Do you know what it means for \phi to be onto? It means that for ANY infinitely differentiable function g we can find a function f such that \phi(f) = g. You have shown that \phi(f) must be 0 when evaluated at 0. Are all infinitely differentiable functions 0 at 0? If you can find some function which is not, then you have your counterexample.

 

[jeff1evesque]

I know onto means that for every element in the domain, there must exist some corresponding element in the range. I am confused by the directions,

...that have derivatives of all orders.

Does this mean f will have derivatives of all order, or will g have derivatives of all order?

You have shown that \phi(f) must be 0 when evaluated at 0.

Why do I have to show this?

 

[rasmhop]

Does this mean f will have derivatives of all order, or will g have derivatives of all order?

f and g are both in F as \phi is a function from F to F, so both f and g must have derivatives of all orders.

Why do I have to show this?

You don't HAVE to show it, but you mentioned \phi(f)(0) = 0 for all f yourself, which is exactly the same as saying that \phi(f)[/tex] is 0 when evaluated at 0 for all f so I assumed you had shown it and suspected it could be used.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I know onto means that for every element in the domain, there must exist some corresponding element in the range. I am confused by the directions, </div> </div> </blockquote>Maybe this was simply a mixup when you typed in the definition, but that is true for all functions and not really the definition of onto. A function f : A \to B is said to be onto if for all elements b of B we can find some element a of A such that f(a) =b. For your problem it means that for all infinitely differentiable functions g (remember this is the group we are considering) we can find an infinitely differentiable function f such that \phi(f) = g. We don&#039;t know what g is as it&#039;s arbitrary but we have:<br /> g(x) = \int_0^x f(t)dt<br /> which suggests trying to substitute x=0.<br /> g(0) = \int_0^0 f(t) dt = 0<br /> Thus we know that for all infinitely differentiable functions g we have g(0)=0, but is this really true? If not then we have a contradiction so \phi can&#039;t be onto. Can you find an infinitely differentiable function that is not 0 at 0? (A very simple example will do)

 

[jeff1evesque]

How about we let the function g = 2t + 1.
Therefore,
<br /> g(0) = \int_0^0 f(t) dt = \int_0^0 [2t + 1] dt = t^{2} + t = 0<br />

But it's suppose to not equal 0, so we can produce the proposed contradiction. I don't know why but this is not obvious to me, I can't find the contradiction.

 

[rasmhop]

How about we let the function g = 2t + 1.
Therefore,
<br /> g(0) = \int_0^0 f(t) dt = \int_0^0 [2t + 1] dt = t^{2} + t = 0<br />

But it's suppose to not equal 0, so we can produce the proposed contradiction. I don't know why but this is not obvious to me, I can't find the contradiction.

You can't see the contradiction in the following?
- Let g(x) = 2x + 1
- Assume \phi is onto.
- Then there exists some function f such that \phi(f) = g
- We have, as you showed, g(0)=0.
- We also have by plugging in x=0: g(0) = 2*0+1 = 1
- Since g is a function g(0) can't be both 0 and 1 so we have a contradiction.
- Thus \phi cannot be onto.

 

[jeff1evesque]

You can't see the contradiction in the following?
- Let g(x) = 2x + 1
- Assume \phi is onto.
- Then there exists some function f such that \phi(f) = g
- We have, as you showed, g(0)=0.
- We also have by plugging in x=0: g(0) = 2*0+1 = 1
- Since g is a function g(0) can't be both 0 and 1 so we have a contradiction.
- Thus \phi cannot be onto.

Onto (as you defined) simply means for all elements b of B we can find some element a of A such that f(a) = b. In this way of thinking, our b is essentially
<br /> \int_0^x [2t + 1] dt,
and not defined as 2x + 1 (or is it)? So why do we have to substitute zero into both ?