# Determining whether the map is an isomorphism

## Homework Statement

Let F be the set of all functions f mapping R into R that have derivatives of all orders. Determine whether p is an isomorphism of the first binary structure with the second.

1. $$<F, +>$$ with $$<R, +>$$ where p(f) = f'(0)

2. $$<F, +>$$ with $$<F, +>$$ where p(f)(x) = $$\int^{x}_{0} f(t)dt$$

3. $$<F, +>$$ with $$<F, +>$$ where p(f)(x) = $$d/dx \int^{x}_{0} f(t)dt$$

4. $$<F, \cdot>$$ with $$<F, \cdot>$$ where p(f)(x) = $$x \cdot f(x)$$

## The Attempt at a Solution

Some ideas I have:

I think #1 is false, since f=x and f=x+1 can have the same derivative at 0. Isomorphism requires that p be bijective.

#3 is true. Simplifying gives p(f)(x)=f(t) for all f in F and x in R.

I'm not so sure about #2, and #4... The book says they are both false, but I don't really understand it. I'd appreciate any hints/suggestions.

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Office_Shredder
Staff Emeritus
Gold Member
It says isomorphism, not just bijection. So for number 3 you also need to show that it preserves the binary structure, i.e. p(f+g)=p(f)+p(g)

For number 4 I don't understand the notation. What is x(x) supposed to be?

For 2, there are two questions: Is p a bijection, and does p preserve addition? Why don't you try a couple examples for $$f(x)$$ to get a feel for what p is doing

I fixed #4. Sorry about that.

For #3, p(f) = f(t), so p is an identity, right? So p(f+g) = f(t) + g(t) = p(f) + p(g).

Office_Shredder
Staff Emeritus
Gold Member
Ok, so let's look at number 2. There are two questions to answer:

Is p(f+g)=p(f)+p(g)?

And is p(f) a bijection?

p preserves addition, but it's definitely not a bijection... I don't think it's onto. i.e. $$p(f)(x)=\int^{x}_{0} f(t) dt \neq x^2$$ for any f in F even though $$x^2$$ is in R...

Does that make sense?