Why is the image of the pre-image of B, a subset of B?

Question

$f: X \to Y$ prove that if $B \subseteq Y$, then the image of the pre-image of $B$ is a subset of B, shown by these symbols respectively: $f[f^{-1}[A]]$

However, I don't see how this is true, because it was previously proven that: $A \subseteq f^{-1}[f[A]]$ where $A \subseteq X$

If that's true, let $B = f[A]$ and let $A \subseteq f^{-1}[B] = C$, so $f[f^{-1}[B]] \nsubseteq B$ which is obvious because C is a superset of A.

What's the problem?

Answer 1

For simplicity, consider the zero-function $f:\mathbb R\to\{0\}$ that sends everything to zero. The we can construct an example following your suggested contradictory pattern as follows:

• $A=[0,1]$
• $B=f(A)=\{0\}$
• $C=f^{-1}(B)=\mathbb R$

Now clearly $A=[0,1]\subseteq C=\mathbb R$, still $f(C)=\{0\}=B$. There is just no problem there.

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In general, when you define $B=f(A)$ you can be certain that every element of $B$ is in the image of $f$ and therefore has a non-empty pre-image. Hence $f(f^{-1}(B))=B$ and not just $f(f^{-1}(B))\subseteq B$ for all such examples.

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Regarding the actual proof of the proposition, the pre-image is defined as follows: $$ f^{-1}(B)=\{x\in X\ \mid\ f(x)\in B\} $$ so clearly for any $x\in f^{-1}(B)$ we must have $f(x)\in B$. Hence $$ f(f^{-1}(B))\subseteq B $$

Answer 2

If an element of $B$ is not the image of any element of $X$, it will not be in the image of the preimage of $B$.

Answer 3

$f (f^{-1}(B))\subseteq B $, for $x\in f^{-1}(B)\implies f (x)\in B $, by definition

The reverse, $B\subseteq f (f^{-1}(B) $ will be true when $f $ is onto $B $, so that, in this case, $f (f^{-1}(B))=B $...

Answer 4

For any $x$ we have $x\in f^{-1}B \iff f(x)\in B.$ So whenever we write $x\in f^{-1}B$ we can equivalently write $f(x)\in B.$ Therefore $$f(f^{-1}B)=\{f(x): x\in f^{-1}B\}=\{f(x): f(x)\in B\}$$ which is a subset of $B.$