Denote $F =\{f:\Bbb R\to \Bbb R\mid f \text{ is infinitely differentiable}\}$.

Define $\varphi:(F,+)\to(F,+)$ by $[φ(f)](x) = \int_0^x f(t) dt $.

Is this map an isomorphism or not?

Pretty much don't know how to dissect the problem and solve it. Statements which are confusing me further :
a) how are the derivatives of all order of F plays in this problem?
b) Since both group under addition, how can I prove yes or no for using the homomorphism,onto,1-1 ?

Any help greatly appreciated. Thanks

Sorry for not bing clear : Question: Is set F isomorphic to F under that function?

  • $\begingroup$ What are you trying to show? This is very unclear. $\endgroup$ – Pedro Tamaroff Oct 4 '14 at 19:03
  • $\begingroup$ It's hard to tell what it is that you're asking. Are you asking if $\varphi$ is an isomorphism? $\endgroup$ – Omnomnomnom Oct 4 '14 at 19:03
  • 1
    $\begingroup$ The derivatives part just makes it a map from $C^\infty$ into itself. If the domain were $C^k$ instead of $C^\infty$ then it would be a map from $C^k$ into $C^{k+1}$. $\endgroup$ – Ian Oct 4 '14 at 19:29

Hint: Note that $\varphi$ is a homomorphism.

Is it surjective (onto)? Is it injective (one to one)? Since $\varphi$ is a homomorphism, it is only an isomorphism if it is both injective and surjective.

What do we know about $[\varphi(f)](0)$? What does this mean for $\varphi$?

A more thorough answer: $\varphi$ is not surjective, so it can't be an isomorphism. If it's not clear why, feel free to leave a comment below.

A point of mathematical grammar: the word "isomorphic" is not used to describe an mapping (i.e. an isomorphism) but the algebraic sets involved (or groups, in this case).

Two groups are isomorphic if there exists any isomorphism between them. The map from one group to another is not "isomorphic"; it "is an isomorphism".

Furthermore, it makes no sense to say that two groups are "isomorphic under" a function. Two groups are isomorphic because one of the functions between them happens to be an isomorphism. There may be maps between them that are not isomorphisms, but those don't have anything to do with whether the spaces are isomorphic.

So, in summary: an "isomorphism" is a type of function between two groups, and "isomorphic" is an adjective used only to describe pairs of groups (or more general algebraic structures), not a potential "isomorphism" between them.


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