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Let $f:A\to B$ be a function. Let $T_1$ and $T_2$ be subsets of $B$. Show that if $f$ is onto, then $$f^{-1}(T_1)\subset f^{-1}(T_2) \implies T_1\subset T_2$$

I proved it as follows.

Let $x\in f^{-1}(T_1)$

=> $\{x\in f^{-1}(T_1) \implies x\in f^{-1}(T_2)\}$

=> $\{f(x)\in T_1 \implies f(x)\in T_2\}$

If $f$ is not onto, there is $b$ in $T_1$ such that no element in $T_1$ hits $b$. So, not all elements in $T_1$, $T_2$ have inverse image under $f$. But since $f$ is onto, all elements in $B$ have their inverse image under $f$. Thus, we can say that all $f(x)\in T_1$ => all $f(x)\in T_2$. Thus, $T_1\subset T_2$.

I think this is incorrect but don't know what is wrong.

Please help me. Thank you so much for helping me.

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To prove your statement, you should start with an element $x \in T_1$ and show it is in $T_2$. Your proof is sort of alright, but your use of the fact that $f$ is onto is somewhat unprecise. – Hrodelbert Apr 12 '14 at 8:14
Thank you so much. I'll try again! – luvmathematics Apr 12 '14 at 8:16
@Hrodelbert How about this? I proved it again as follows. Let $y$∈$T$1. Since $f$ is onto, all $y$∈$T$1 have their inverse image under $f$. Then there exists $x$∈$f$-1($T$1) s.t $f(x)$=$y$∈$T$1. $x$∈$f$-1($T$1)⊂$f$-1($T$2) => $x$∈$f$-1($T$2) => $f(x)$=$y$∈$T$2 => $y$∈$T$2 Thus, $T$1⊂$T$2. – luvmathematics Apr 12 '14 at 8:33
This is exactly right! Well done! – Hrodelbert Apr 12 '14 at 8:43

Let's try simplifying the reasoning (the one you do in comments is good, but it can be better worded).

Assume $f$ is onto and that $f^{-1}(T_1)\subset f^{-1}(T_2)$. You want to deduce from these assumptions that $T_1\subset T_2$.

You have then to take $b\in T_1$ and prove $b\in T_2$. Since $f$ is onto, there is $a\in A$ such that $f(a)=b$. By definition, $a\in f^{-1}(T_1)$ and so, by assumption, $a\in f^{-1}(T_2)$. By definition of $f^{-1}$ this means $f(a)\in T_2$, so $b\in T_2$.

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