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

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

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

(b) $\varphi$ surjective -> $T$ 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 $\mathbb{R}^A=\{f\text{ such that }f: A \rightarrow \mathbb{R}\}$ (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 $\phi: A \rightarrow B$. Suppose $\varphi$ is injective. Let $g:A\rightarrow \mathbb{R}$ be any function.
We need to find a function $f: B\rightarrow \mathbb{R}$ so that $g=f\circ \varphi$.
To do this, first note that since $\varphi$ is injective, it is actually a bijection $A\rightarrow \text{Range}(\varphi)$,
so there exists $\gamma:\text{Range}(\varphi)\rightarrow A$ such that $\gamma \circ \varphi(a)=a$ for all $a\in A$.
Then define $f(b)=0$ if $b\notin \text{Range}(\varphi)$ and $f(b)=g(\gamma(b))$ otherwise. Then $f \circ \phi (a)=f(\phi(a))=g(\gamma(\phi(a)))=g(a)$. 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 $\mathbb{R}^A$?

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

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

$T(f) = T(g)$ means that $f \circ \varphi = g \circ \varphi$, so that f and g are equal for all elements in $B$ that are in the range of $\varphi$. If $\varphi$ is surjective, then the range of $\varphi$ is all of $B$ and so $f$ and $g$ 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 $\mathcal{F}=\{ f:A\rightarrow B\text{ such that } f\text{ satisfies...}\}$.
It may be less "slick" but it sure prevents any and all confusion about what you are saying.