Let $f:A \rightarrow B$ be a map and let $X \subseteq A$ and $Y \subseteq B$ Then by definition, we have the following two sets:
The image of $X$, which is the set $f(X)=$ { $b\in B$ | $b = f(x)$ for some $x \in X$}.
The inverse image of $Y$, which is the set $f^{-1}(Y)=$ { $a \in A$ | $f(a) \in Y$}.
I read that we can think of the map $f$ as inducing the maps $f_*: \mathcal{P}(A) \rightarrow \mathcal{P}(B)$ and $f^*: \mathcal{P}(B) \rightarrow \mathcal{P}(A)$. But I’m not really seeing how I can use this maps.
My doubt arose from the following problem (from a book about fundamentals) which I’m still working on but (yet) without success:
Let $f,g: A \rightarrow B$ be maps. Think of these maps as inducing maps $f_*,g_*: \mathcal{P}(A) \rightarrow \mathcal{P}(B)$, and maps $f^*,g^*: \mathcal{P}(B) \rightarrow \mathcal{P}(A)$. Show that $f_* = g_*$ iff $f^* = g^*$ iff $f=g$.
I would really appreciate if someone could:
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Explain what these maps $f_*,g_*$ and $f^*,g^*$ are, what is the “motivation”/relation between the original maps $f,g$ And how do they relate with the concepts of image and inverse image.
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I’m not yet looking for the solution for this problem, but still I would like to get some advices about “what should I do” or “where should I start”.
Thank you for your attention!
Best Answer
If $f=g$ and $X\subset A$, then$$f_*(X)=\{f(x)\mid x\in X\}=\{g(x)\mid x\in X\}=g_*(X).$$And if $f\ne g$, take some $a\in A$ such that $f(a)\ne g(a)$. Then $f_*(\{a\})\ne g_*(\{a\})$, since $f_*(\{a\})=\{f(a)\}$ and $g_*(\{a\})=\{g(a)\}$.
So, I have proved that $f=g\iff f_*=g_*$.
Can you prove now that $f=g\iff f^*=g^*$?