# Thread: Linear transformation between function spaces

1. ## Linear transformation between function spaces

Hi all.

I am a beginner to real maths and am looking at a problem from a book by Loomis and Sternberg.

Let $\displaystyle T$ be the linear transformation from $\displaystyle \mathbb{R}^B$ to $\displaystyle \mathbb{R}^A$ such that $\displaystyle T(f) = f \circ \varphi$.

I need to show that $\displaystyle T$ is an isomorphism if $\displaystyle \varphi$ is a bijection by showing that:

(a) $\displaystyle \varphi$ injective -> $\displaystyle T$ surjective;

(b) $\displaystyle \varphi$ surjective -> $\displaystyle T$ injective.

I'm not really sure where to start, and I can't even see the link between the properties of the 2 mappings; any hints would be greatly appreciated.

Thanks

2. Is $\displaystyle \mathbb{R}^A=\{f\text{ such that }f: A \rightarrow \mathbb{R}\}$ (I hate this notation), and in what sense do you mean isomorphism?

If I am right on the notation, I suppose it must be that $\displaystyle \phi: A \rightarrow B$. Suppose $\displaystyle \varphi$ is injective. Let $\displaystyle g:A\rightarrow \mathbb{R}$ be any function.
We need to find a function $\displaystyle f: B\rightarrow \mathbb{R}$ so that $\displaystyle g=f\circ \varphi$.
To do this, first note that since $\displaystyle \varphi$ is injective, it is actually a bijection $\displaystyle A\rightarrow \text{Range}(\varphi)$,
so there exists $\displaystyle \gamma:\text{Range}(\varphi)\rightarrow A$ such that $\displaystyle \gamma \circ \varphi(a)=a$ for all $\displaystyle a\in A$.
Then define $\displaystyle f(b)=0$ if $\displaystyle b\notin \text{Range}(\varphi)$ and $\displaystyle f(b)=g(\gamma(b))$ otherwise. Then $\displaystyle f \circ \phi (a)=f(\phi(a))=g(\gamma(\phi(a)))=g(a)$. Since we have found a preimage for an arbitrary element, the function is onto.

See if you can do (b) on your own now.

3. Thanks for that, I was on completely the wrong lines - I was thinking it was something to do with the linearity. You were right about the notation. Out of interest, what do you use to represent spaces like $\displaystyle \mathbb{R}^A$?

As for (b), if I'm right then it's easier than (a):

Let $\displaystyle f$ and $\displaystyle g$ be functions from $\displaystyle B$ to $\displaystyle \mathbb{R}$. We need to show that $\displaystyle T(f) = T(g) \Rightarrow f = g$, then T is injective.

$\displaystyle T(f) = T(g)$ means that $\displaystyle f \circ \varphi = g \circ \varphi$, so that f and g are equal for all elements in $\displaystyle B$ that are in the range of $\displaystyle \varphi$. If $\displaystyle \varphi$ is surjective, then the range of $\displaystyle \varphi$ is all of $\displaystyle B$ and so $\displaystyle f$ and $\displaystyle g$ are equal everywhere in their domain.

Does this seem right? Thanks for your help!

4. That looks right to me!

As for the notation, I just define a set to be all such functions, i.e. let $\displaystyle \mathcal{F}=\{ f:A\rightarrow B\text{ such that } f\text{ satisfies...}\}$.
It may be less "slick" but it sure prevents any and all confusion about what you are saying.