Is Composition by a Mapping a Linear Isomorphism in Vector Spaces?

[fluxions]

Homework Statement

Sorry for the vague title!

Let R denote the set of real numbers, and F(S,R) denote the set of all functions from a set S to R.

Part 1: Let $$\phi$$ be any mapping from a set A to a set B. Show that composition by $$\phi$$ is a linear mapping from F(B,R) to F(A,R). That is, show that $$T : F(B,R) \rightarrow F(A,R) : f \mapsto f \circ \phi$$ is linear.

Part 2: In the situation given in part 1, show that T is an isomorphism if $$\phi$$ is bijective by show that:
(a) $$\phi$$ injective implies T surjective;
(b) $$\phi$$ surjective implies T injective.

The Attempt at a Solution

Well, I got part 1.
As for part 2... I have no clue. Any ideas?

 

[LCKurtz]

OK, I will get you started on (a.) So we are supposing $\phi$ is 1-1. So for each a in A there is a unique b in B such that $b = \phi(a)$ and $a =\phi^{-1}(b)$. Now suppose you are given g in F(A,R). To show T is onto, you need to build an f in F(B,R) such that T(f) = g, that is $f\circ\phi = g$.

For $b \in \phi(A)$, try defining $f(b) = g(\phi^{-1}(b))$. Now see if you can show that T(f) = g, which is the same as showing $f\circ\phi = g$.

Also note, if b is exterior to $\phi(A)$, it doesn't matter what you define f(b).