# Dual space topology

1. Oct 7, 2006

### kakarukeys

let
$$A \hookrightarrow B$$
be a continuous inclusion map from A to B.
A, B are two topological spaces. $$A \subset B$$

what can we say about the induced map between topological dual spaces
$$B^* \hookrightarrow A^*$$?

is it continuous and injective?

2. Oct 7, 2006

### Hurkyl

Staff Emeritus
Look at special cases. If we have an injective map

$$\mathbb{R}^2 \rightarrow \mathbb{R}^3$$

then when we dualize, what sort of map do we get on the dual spaces:

$$\mathbb{R}^2^* \cong \mathbb{R}^2 \leftarrow \mathbb{R}^3 \cong \mathbb{R}^3^*$$

?

Dualities tend to reverse most notions. "monic" and "epic" are dual notions, so when you dualize a monomorphism, it tends to become an epimorphism, and vice versa.

Of course, it's always a good idea to work out the details for yourself. It gives you good practice with the notions involved.

3. Oct 7, 2006

### kakarukeys

thanks...
But it doesn't solve my problem of understanding certain steps in
http://en.wikipedia.org/wiki/Rigged_hilbert_space

"that is one for which the natural inclusion
$$\Phi\subset H$$
is continuous. It is no loss to assume that $$\Phi$$ is dense in H for the Hilbert norm. We consider the inclusion of dual spaces $$H^*$$ in $$\Phi^*$$."

Why $$\Phi\subset H \implies H^*\subset\Phi^*$$?
Could you please take a look?

4. Oct 8, 2006

### Hurkyl

Staff Emeritus
I'm confused. You just said you understand why $A \longmapsto B$ implies $B^* \longmapsto A^*$. So, I don't see why you don't get $\Phi \longmapsto H$ implies $H^* \longmapsto \Phi^*$.

5. Oct 8, 2006

### kakarukeys

no... in my second post, I used set operation "$$\subset$$"
I understand there is a map from $$H^*$$ to [/tex]]Phi^*[/tex] which is the induced map. The wiki articles says
$$\Phi\subset H \implies H^*\subset\Phi^*$$

H* a subset of Phi* ??? in what sense?
I mean the functions in H* and the functions in Phi* have different domains, how could the two be related?

6. Oct 8, 2006

### matt grime

Set inclusion is a function. Duality reverses arrows, and hence inclusions. Also very few topoological spaces have any notion of duality.

7. Oct 8, 2006

### Hurkyl

Staff Emeritus
Ah, so that's what you're worried about.

Generally, injective maps are what matter, whether or not you have an actual subset is irrelevant. But inclusions are notationally convenient... so sometimes (for convenience) when we have an injective map, we identify the the objects of the domain with their images.

Basically, because the restriction map $H^* \longmapsto \Phi^*$ is so natural, there isn't really any benefit in making a distinction between an element of $H^*$ and its image.

8. Oct 8, 2006

### kakarukeys

so is it right to say the wiki article is wrong?
the map from $$H^*$$ to $$\Phi^*$$ is onto but not injective.
So there is no inclusion and $$H^*$$ is not a subset of $$\Phi^*$$.

9. Oct 9, 2006

### Hurkyl

Staff Emeritus
Hrm. Now that I think more about it, I think the maps $\Phi \rightarrow H$ and $H^* \rightarrow \Phi^*$ are both monic and epic in the categorical sense. I know that epic doesn't always imply surjective... I don't know if monic implies injective in this category. (For a simpler example of what I mean, the ring homomorphism $\mathbb{Z} \rightarrow \mathbb{Q}$ is epic... but it's not surjective)

Oh well; if we want to prove injectivity, that just means we'll have to resort to a dirtier method. I think it's not too difficult to prove that the kernel of the map $H^* \rightarrow \Phi^*$ is zero.

10. Oct 9, 2006

### kakarukeys

the map $$H^* \longmapsto \Phi^*$$ is injective and not surjective.
it's injective because $$\Phi$$ is dense in H, you can prove injectivity in a few steps.