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$f:X \rightarrow Y$ where $A \subset X$ and $B \subset Y$

Prove (if it is correct) that $$B \subset f(f^{-1}(B))$$ This is my approach

Proof

Choose an arbitrary element $y \in B$, by the definition of inverse image, we know that $x \in f^{-1}(B)$ such that $f(x)=y$. As $y \in B$ and $x \in f^{-1}(B)$, we can conclude that $y \in f(f^{-1}(B))$.

Is this correct?

NimaJan
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  • Your mistake: $y\in B$ does not ensure the existence of an $x$ with $f(x)=y$. This because $f$ is not necessarily surjective so that it might happen that $y$ is not in the range of $f$. – drhab Jan 08 '20 at 13:01

2 Answers2

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This is false. Let $A = X = \{0\}$, $Y = \{0, 1\}$ and $B = \{1\}$. Let $f : X \to Y$ be the identity function. Then $f^{-1}(B) = \emptyset$, so $f(f^{-1}(B)) = \emptyset$ but $B$ is not empty.

The error in your reasoning is the statement "...by the definition of inverse image, we know that $x \in f^{-1}(B)$ such that $f(x) = y$". This is not true because the definition of inverse image states that: $$ f^{-1}(B) = \{x \in X \mid f(x) \in B\} $$ This definition asserts that $f$ maps every element in $f^{-1}(B)$ to $B$, but not necessarily to every element in $B$.

As such, if you assume that $f$ is surjective, then your statement holds - you can safely conclude that such an $x$ which $f(x) = y$ exists.

Clement Yung
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  • Thank you for your clean explanation. Is this counterexample I made, valid? https://imgur.com/2SZs7IH – NimaJan Jan 08 '20 at 11:58
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    That's more of an illustration of the idea to why the statement is false than a concrete counterexample, but your idea in the illustration is correct. – Clement Yung Jan 08 '20 at 12:00
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The statement you provide is in general wrong. It is the other inclusion that always holds.

J. De Ro
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