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General vector space question

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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 [tex]\phi[/tex] be any mapping from a set A to a set B. Show that composition by [tex]\phi[/tex] is a linear mapping from F(B,R) to F(A,R). That is, show that [tex] T : F(B,R) \rightarrow F(A,R) : f \mapsto f \circ \phi[/tex] is linear.

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



The Attempt at a Solution


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

Answers and Replies

  • #2
LCKurtz
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OK, I will get you started on (a.) So we are supposing [itex]\phi[/itex] is 1-1. So for each a in A there is a unique b in B such that [itex]b = \phi(a)[/itex] and [itex]a =\phi^{-1}(b)[/itex]. 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 [itex]f\circ\phi = g[/itex].

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

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

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