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Suppose $F$ is the set of all functions $f$ mapping $\mathbb{R}$ into $\mathbb{R}$ that have derivatives of all order. Then is $\phi$ an isomorphism of the first binary operation with the second?

  1. $<F,+>$ with $<F,+>$ where $\phi(f) = \int_{0}^{x}f'(t) dt$

No, because $\phi$ does not map $F$ onto $F$. For all $f\in F$, we see that $\phi(f)(0)=0$ so, for example, no function is mapped by $\phi$ into $x+1$.

I am not sure what the solution means.

Is it saying that if $x = 0$, then $\int_{0}^{0}f(t)dt = 0 \implies \phi(x+1)(0) = \int_{0}^{0}t+1 dt = 0$? Because that is true. So why isn't it onto?

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The solution is saying that for any $f$, $\phi(f)(0) = \int_0^0 f'(t) d(t) = 0$ but set $g = x+1$ then $g(0) = 1$, so there can be no $f$ such that $\phi(f) = g$. – Deven Ware Oct 12 '12 at 4:34
What's "B.O"? $ $ – joriki Oct 12 '12 at 5:26
Binary Operation – Hawk Oct 12 '12 at 5:53
Quite generally it's a good idea to introduce any abbreviations that you use. That's especially relevant when your use of them is flawed, since that makes it even harder to guess them. An isomorphism is between structures, e.g. groups, not between operations. (Note that there's an edit button under the question.) – joriki Oct 12 '12 at 8:15
@jak This unfortunate choice of abbreviation is doubly unfortunate in English, where "B.O." is "body odor"... – rschwieb Oct 12 '12 at 12:39
up vote 2 down vote accepted

For every function in $\mathrm{Im}(\phi)$, the image of $0$ is $0$.

There are many functions which are infinitely differentiable which are not $0$ when evaluated at $0$.

Therefore $\mathrm{Im}(\phi)\neq F$.

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