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I need help with the following problem: given the function $f:A \rightarrow B$ with sets $A_0 \subset A$ and $B_0 \subset B$, I need to prove that $f(f^{-1}(B_0)) \subset B_0$ and that equality holds only when $f$ is surjective.

I started with the set $f^{-1}(B_0)$, take $b\in B_0$ I considered three scenarios:

1.- there is just one $a \in A$ such that $f(a)=b$

2.- there is more than one element in $A$, say $a_1, a_2, \dots, a_n \in A$ such that $f(a_i)=b$, $i=1,2, \dots,n$

3.- there is no element $a \in A$ such that $f(a)=b$

So, if we compose $f$ with $f^{-1}(b_0)$, $f(f^{-1}(B_0))$, from points 1 and 2 we'll get back an element $b \in B_0$, but for point 3 $B_0$ might have more elements, $\forall x \in f(f^{-1}(B_0)), x\in B_0$ but $\exists b \in B_0, b \notin f(f^{-1}(B_0))$, hence, $f(f^{-1}(B_0)) \subset B0$.

If $f$ is surjective then there is a guarantee that every $b \in B_0$ will have at least an $a \in A$ such that $f(a)=b$, meaning $f(f^{-1}(B_0)) = B0$

I was told that this is wrong, but I don't have a clue of where the procedure failed, any pointers will be appreciated.

Asaf Karagila
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Jacob Morales Gonzalez
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  • It is not clear what exactly you were told that this is wrong, you have written a lot. In general for the proof you have to start with $y \in f(f^{-1}(B_0)) $. – zkutch Jun 21 '23 at 12:29
  • For more info and similar problems https://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages – Popocatepetl112358 Jun 21 '23 at 16:50
  • More formal, then mentioned link https://math.stackexchange.com/questions/4721056/mathematical-analysis-i-by-zorich-show-f-1fa-supset-a-ff-1b/4721073#4721073 – zkutch Jun 21 '23 at 17:03

1 Answers1

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Having definitions for $A_0\subset A, B_0\subset B$ $$f(A_0)=\{ y: (\exists x)(x\in A_0 \land y=f(x)) \} \\ f^{-1}(B_0) = \{x:f(x)\in B_0\} $$ Let's start, as I suggest in comment: $$y \in f(f^{-1}(B_0)) \Leftrightarrow (\exists x)\Big(x\in f^{−1}(B_0) \land y=f(x)\Big) \Leftrightarrow \\ \Leftrightarrow (\exists x)\Big(f(x) \in B_0 \land y=f(x)\Big) \Rightarrow y \in B_0$$,

Now, let's assume surjectivity, i.e. $f(A)=B$, and start reverse direction: $$y \in B_0 \subset B \Rightarrow (\exists x)\Big(x\in A \land y=f(x)\in B_0 \subset B \Big) \Rightarrow \\ \Rightarrow (\exists x)\Big(x\in f^{−1}(B_0) \land y=f(x)\in B_0 \Big) \Rightarrow y=f(x)\in f(f^{−1}(B_0))$$

zkutch
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