Loomis and Sternberg Problem 1.46

I have no clue about how to solve 1.46(a). Here is my attempt at 1.46(b):

Let $f_0^A$ be the additive identity of $\mathbb{R}^A$ and let $f_0^B$ be the additive identity of $\mathbb{R}^B$. The linear map $T$ is injective if its null space is $\{f_0^B\}$.

$f_0^B$ is in the null space of $T$ since $T(f_0^B)=f_0^B \circ \varphi = f_0^A$.

Only $f_0^B$ is in the null space because if $T(\theta)= \theta \circ \varphi = f_0^A$, then $\theta $ has to be a function that maps every element of $B$ to $0 \in \mathbb{R}$ since $A$ is surjective (not sure how to make this argument more formal). Thus $\theta = f_0^B $ and the null space of $ T$ is $\{f_0^B\}$.

Any hints are appreciated. Thanks!

Solution 1:

For 1.46 a) take any $g\in \mathbb R^{A}$. This means $g$ is a map from $A$ to $\mathbb R$. Define $f: B \to \mathbb R$ by $f(b)=g(a)$ if $b$ is of the form $\phi (a)$ for some $a \in A$ (which means $b$ is in the range of $\phi$). Otherwise define $f(b)=0$. If $b=f(a)=f(a')$ the $a=a'$ (because $\phi$ is injective) and hence $g(b)$ is well-defined. Now $f\in \mathbb R^{B}$ and $T(f)=g$ because $f\circ \phi =g$.

1.46 b) is much simpler. Suppse $Tf=Tf'$. Then $f(\phi (a))=f'(\phi (a))$ for all $a \in A$. Since $\phi$ is surjective this implies $f(b)=f'(b)$ for all $b \in B$ so $f=f'$.