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I have no idea, how to solve this problem:

Let $\psi: (\mathbb{R} \to \mathbb{N}) \to P(P(\mathbb{R}))$ defined as:

$\psi (f) = \mathbb{R}/_{ker(f)}$

(1) Is this a surjective function?

(2)Is this an injective function?

I tried to do something, but after hours I can't see any progress... I just don't know, were/how to start solving such problem. And I'm not sure, if I properly understand problem itself; I'm sure, that your explanation woluld be helpful.

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    You seem to have misstated your question, for example $f$ is probably a homomorphism, and I can't see why the image should be $P(P(N))$.2011-01-02
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    As for (2), consider $f$ vs. $2f$.2011-01-02
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    As for (1), the idea is probably that the image of $\psi$ is not arbitrary, but a proper statement of the question is required here.2011-01-02
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    @Yuval Filmus: Yep, I made a mistake, there should be $P(P(R))$, not $P(P(N))$. Well, what now?2011-01-02
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    You should explain your notation. What is P(P(R))? Whats ker f (this could be answered by saying what type of maps you're looking at from R to N)? Whats R/ker f? Is this a quotient of groups?2011-01-02
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    Is $R/\ker f$ the set of all conjugacy classes? This would be a member of $P(P(R))$. It will only make sense if $f$ is a homomorphism.2011-01-02

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