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Problem 1. Let $f \colon X\to Y$ be a function from the set $X$ to the set $Y$. For a subset $A\subset X$, let $f_*(A)=\{ y\in Y | \exists x\in A\text{ such that }f(x)=y\}$. Prove the following:
(i) The function $f \colon X\to Y$ is one-to-one if and only if $f_*(A\cap B)=f_*(A) \cap f_*(B)$ for all subsets $A$, $B$ of $X$.
(ii) The function $f \colon X\to Y$ is onto if and only if $Y\setminus f_*(A) \subseteq f_*(X\setminus A)$ for all subsets $A$ of $X$.

I've managed to prove the first part of problem 1. Stuck on the second. And completely clueless about problem 2.

My attempt: Problem 1
i) Suppose $y\in f_*(A \cap B)$. Then $f_*(x)=y$ such that $x \in A \cap B$. Then, $x \in A$ and $x \in B$. Therefore, $y \in f_*(A)$ and $y \in f_*(B)$.
This implies $y \in f_*(A) \cap f_*(B)$.
Therefore, $f_*(A\cap B)=f_*(A) \cap f_*(B)$.

Martin Sleziak
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Amy Sandhu
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    Can we see some work? – Julian Rachman Feb 02 '15 at 06:25
  • http://i.stack.imgur.com/BTLmO.jpg – Amy Sandhu Feb 02 '15 at 06:53
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    Problem 1(ii) might make a decent question, but you really should include the work you did in the question itself rather than a link from a comment--and even better, typeset all this so it's readable. Problem 2 seems unrelated. I'm not sure this is even the right site for it. The two problems certainly should not be combined in one question. – David K Feb 02 '15 at 06:54
  • Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Feb 02 '15 at 08:51
  • Maybe you can find answer to at least some of your questions here: http://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages – Martin Sleziak Feb 02 '15 at 08:56
  • For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Feb 02 '15 at 09:04

1 Answers1

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In your attempt you prove that $f_{*}\left(A\cap B\right)\subseteq f_{*}\left(A\right)\cap f_{*}\left(B\right)$ but not that $f_{*}\left(A\cap B\right)=f_{*}\left(A\right)\cap f_{*}\left(B\right)$ as you claim.

(i)$\Rightarrow$ From $A\cap B\subseteq A\wedge A\cap B\subseteq B$ it follows immediately that $f_{*}\left(A\cap B\right)\subseteq f_{*}\left(A\right)\cap f_{*}\left(B\right)$ (as you show in your attempt). If $y\in f_{*}\left(A\right)\cap f_{*}\left(B\right)$ then $y=f\left(a\right)=f\left(b\right)$ for some $a\in A$ and some $b\in B$. Then $a=b\in A\cap B$ (because $f$ is one-to-one) so that $y\in f_{*}\left(A\cap B\right)$.

(i) $\Leftarrow$ Let $u,v\in X$ with $y=f\left(u\right)=f\left(v\right)$. For $A=\left\{ u\right\} $ and $B=\left\{ v\right\} $ we find $y\in f_{*}\left(A\right)\cap f_{*}\left(B\right)=f_{*}\left(A\cap B\right)$. This implies that $A\cap B\neq\emptyset$ or equivalently that $u=v$.

(ii) $\Rightarrow$ Let $y\in Y-f_{*}\left(A\right)$. Then $y=f\left(x\right)$ for some $x\in X$. However, if $x\in A$ then $y=f\left(x\right)\in f_{*}\left(A\right)$ contradicting that $y\in Y\backslash f_{*}\left(A\right)$. We conclude that $x\in X-A$ and consequently $y=f\left(x\right)\in f_{*}\left(X-A\right)$.

(ii) $\Leftarrow$Substituting $A=X$ leads to $Y-f_{*}\left(X\right)\subseteq f_{*}\left(X-X\right)=f_{*}\left(\emptyset\right)=\emptyset$. So $Y=f_{*}\left(X\right)$, i.e. $f$ is onto.

drhab
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