show a conclusion from the homomorphism $\phi : \Bbb R _{>0} \to \Bbb R$ such that $\phi (r) = \log(r)$

Question

I need two show homomorphism and get a conclusion from iso1 in the following:

a) I have $\phi : \Bbb R _{>0} \to \Bbb R$

$\phi (r) = \log(r)$

I assume here that $\Bbb R _{>0}$ is with multiplication and $\Bbb R$ is with addition?? Is that correct to assume ?

because I don't have this information in the question. if it is with addition then I don't know how to make homomorphism.

so i get $\ker (\phi) = 1 = e$

so my conclusion from iso1 is that

$\Bbb R _{>0} \cong \Bbb R$

which dosen't make sense to me..

b) I have $\phi : \Bbb R^* \to \Bbb R$

$\phi (r) = log(|r|)$

again I assume that $\Bbb R$ is with addition

and I get to the conclusion that

$\Bbb R^*/Sign \cong \Bbb R$

which again dosen't make sense.

any help will be very appreciated

Answer 1

You are correct in your assumptions--after all, $\Bbb R_{>0}$ isn't a group under addition, and $\Bbb R$ isn't a group under multiplication.

The thing to keep in mind is that, unlike what we see in finite sets, infinite sets can be in one-to-one correspondence with proper subsets of themselves! In fact $\Bbb R$ can be put in one-to-one correspondence with $(a,b)$ for any two $a,b\in\Bbb R$ with $a<b,$ but that is not particularly relevant.

You have in fact proved that $r\mapsto\log(r)$ is a one-to-one homomorphism $\Bbb R_{>0}\to\Bbb R.$ If you can prove it is surjective, then you have proved that it is actually a bijection and so an isomorphism.

From there, since $\Bbb R^*/\{1,-1\}\cong\Bbb R_{>0}$ readily, then the first result gives you another way to verify that $\Bbb R^*/\{1,-1\}\cong\Bbb R$.

Answer 2

They ask you to show there is a homomorphism in both cases. It is left for you to make the additional necessary assumptions to do so. Your guess for this is correct, as $\log(xy)=log(x)+log(y)$

As for your surprise about the correct conclusion that you draw, remember you are dealing with the continuum: the whole real line is contained within any connected open subset of $\mathbb{R}$. For example, $f(x)=\arctan(x)$ is a bijection between $(-\pi/2,\pi/2)$ and $\mathbb{R}$.

Mind you, a similar bijection occur for, say the natural numbers, where there are as many even natural numbers as the whole set of natural numbers.

Answer 3

Yes, your assumptions are correct.

To show that $\phi$ is a homomorphism you must show: $$\phi(xy)=\phi(x)\phi(y)$$

Define: $\phi(x)=log(x), s.t. \phi: \mathbb{R_{>0}} \to \mathbb{R}$

Consider:

$$\phi(xy)=log(xy)=log(x)+log(y)=\phi(x)+\phi(y)$$

So we have that $\phi(x)$ is a homomorphism.

Here, $ker(\phi(x))=1$ since $log(1)=0$ where 1,0 are identity elements of the domain and codomain respectively. So you have that the identity of your domain maps to the identity of your codomain.

Thus, $\phi(x)$ mapping between these two groups as defined is an isomorphism and these two groups are isomorphic.

So, why is your observation true? Infinite sets can map 1-1 onto subsets of themselves. Cantor used this to describe the different sizes of infinite sets to show that the cardinality of (0,1) is the same as the cardinality of $\mathbb{R}$.

Hope this helps a bit.