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$f: X \longrightarrow Y$ and $B \subseteq Y$

I am currentry trying to give an example where $f(f^{-1}(B)) \neq B$.

I know, that for a function to have an inverse, it must be bijective.

My thought process is, that the answer can only be equal to $B$ if the function $f$ is bijective, because only then can every element in $f^{-1}(B)$ be directly traced back to B.

Here's where I struggle: How can $f(f^{-1}(B)) \neq B$ if the function needs to be bijective in the first place? Wouldn't this be a contradiction? Or is that exactly what I need to disprove here?

Asaf Karagila
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schrattinger
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    The function does not need to be bijective! You take the inverse image, this can always be done for any function. – kade Jan 04 '21 at 12:47
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    Take $f: \mathbb{R} \to \mathbb{R}, \quad x \mapsto x^2$. Then $f^{-1}(\mathbb{R}) = \mathbb{R}$ but $f(\mathbb{R}) = [0, \infty)$. – kade Jan 04 '21 at 12:51
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    It is a profoundly crappy notation. My professors used $f^<$ to represent the preimage (and $f_>$ to represent the forward image) and everyone should switch to that immediately. –  Jan 04 '21 at 12:52
  • You can always go with another common alternative, which is $f[B]$. – Prime Mover Jan 04 '21 at 13:00
  • Given an arbitrary relation $S$ over $A$ and $B$, one can define the "inverse relation" $T = {(a,b)~:~(b,a)\in S}$ over $B$ and $A$. This relation is often given the notation $S^{-1}$. This can be done regardless of whether or not $S$ is a function or $S^{-1}$ is a function. There is special significance when $S$ and $S^{-1}$ are both functions, the relation would then necessarily be a bijective function. That is not required here however. – JMoravitz Jan 04 '21 at 13:01

2 Answers2

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The inverse of a mapping may not necessarily be a mapping. $f^{-1} [B]$ (the notation here is superior) is merely "the set of all elements of $X$ whose image under $f$ is in $B$".

$f^{-1} [B] = \{x \in X: f(x) \in B\}$

So $f [f^{-1} [B]]$ is the intersection of $B$ with the image of $f$.

Hence if you have elements of $B$ which are not mapped to by $f$, then you have $f [f^{-1} [B]] \ne B$.

Prime Mover
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For the part where you struggled, it is true that for a function to have an inverse, it must be bijective. But $f^{-1}(B)$ is a subset of $X$ that is called the inverse image of B. It is defined for all function $f$ as $$ f^{-1} (B) = \left\{ x \in X \ | \ f(x) \in B \right\}. $$

for $A \subseteq X$, $f(A)$ is also called the image of A and defined as $$ f(A) = \left\{ f(a) \ | \ a \in A \right\}. $$

Back to the question. For every function, $f(f^{-1}(B)) \subseteq B$ is always true. You can prove this by using two definitions above.

However, $B \subseteq f(f^{-1}(B))$ is false if $f$ is not surjective. The reason is that not every element of $Y$ has an inverse-image. That particular element will be inside $B$, but not $f(f^{-1}(B))$.

For example, let $f(x) = x$, $X = [0,1]$, and $Y = [0,2]$. Let $B = Y$. We have $f^{-1}(B) = [0,1]$, so that $f(f^{-1}(B)) = f([0,1]) = [0,1]$. There, $B \nsubseteq f(f^{-1}(B))$.

CHR
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