let $f: P(\Bbb Z) \to P(\Bbb N)$ definined as $f(X)= \{2x^2+5:x \in X\}$. What is the set $f[f^{-1}[\{A\}]]$, where $A=\{3,4,5\}$?
My thoughts: $f^{-1}(Y)= \left\{ \pm\sqrt{\frac{y-5}2} :y\in \Bbb Y \right\} $, so $f^{-1}(A)= \{ 0\}$. Thus, $\ f[f^{-1}[\{A\}]] = \{5\}$.
Is my reasoning wrong?
Thank you for your help.
If $B = f^{-1}(A)$ then $f(B) \subseteq A$. So we need to solve: $2b^2 + 5 = 3, 4, 5$. Of these, only the last one has integer solution $b = 0$. Thus: $B = \{0\}$, and therefore $f(f^{-1}(A)) = f(B) = f(\{0\}) = \{5\}$. So you are right.
I would be careful with how you defined $f^{-1}(Y)$, note if the set you were concerned with was $B=\{7,13\}$ then your definition would have that $f^{-1}(Y) = \{1, 2\}$ but really it should be $f^{-1}(Y) = \{-2,-1,1,2\}$. Note that the definition, in qualitative terms, is the set of $x$ so that you can send $x$ to some element of $B$. So in qualitative terms you need to think about what integers can be sent to $7,13$ which is exactly what I have above. Using the purely quantitive method to solve this problem we can see that $$ x \to 2x^2 + 5 \implies f^{-1}(Y) = \{ x \in X \mid 2x^2 + 5 \in Y\} $$ which comes down to solving, for each $y \in Y$, the equation $$ y = 2x^2 + 5 $$ for $x$ where $x \in \mathbb{Z}$. I'm sure you see the steps to solve this but I'll list them out anyways for latex practice in the least ;P $$ y = 2x^2 + 5 \implies \frac{y - 5}{2} = x^2 \implies x = {\color{red} \pm} \sqrt{\frac{y-5}{2}} $$ since $x^2 = a$ can be solved by either $a$ or $\color{red}-a$. So applying this to the $B$ I have above we are going to find the exact set I had found.
Now to talk about your problem specifically, since $A = \{3,4,5\}$ we have that $$ f^{-1} (A) = \{0\} $$ since the only solution to the above equation iterating over all elements of $A$ is $x=0$ thus we then look at the image of $\{0\}$ under $f$ which is exactly what you found, $\{5\}$.
