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Suppose $f:R^n→R^m$, where $S,T⊆R^n$, and $A,B⊆R^m$. Determine if the statements are always true, never true, or sometimes true and sometimes false.

$1.f(S∩T)=f(S)∩f(T)$

$2. f^{−1}(A∩B)=f^{−1}(A)∩f^{−1}(B)$

$3. A⊆B⇒f^{−1}(A)⊆f^{−1}(B)$

$4. f(f^{−1}(A))=A$


I think $1$ and $2$ are similar statements which might be sometime true, sometime false, and $3$ might also depends on what $f$ is, $4$ looks fine to me, should be always True, but the answer says that's not right.

Any help or hint or suggestion would be appreciated.

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  • $\begingroup$ It's good to have an intuition about these things, but could you supply a proof if asked to? Or is that what you're asking for? $\endgroup$ – Charles Hudgins Sep 28 '19 at 3:05
  • $\begingroup$ This question not asking for proof, but it would be nice to have one $\endgroup$ – Manx Sep 28 '19 at 3:08
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Your intuition on 1 is correct. Can you think of a counterexample? Hint: consider $f(x) = x^2$ and choose $S$ and $T$ carefully. What property could $f$ have that would guarantee equality?

Note that 2 is always true. This is checked directly. I'll do the first half: $$ x \in f^{-1} (A) \cap f^{-1} (B) \implies f(x) \in A \land f(x) \in B \implies f(x) \in A \cap B \implies x \in f^{-1}(A \cap B) $$ So $f^{-1}(A) \cap f^{-1}(B) \subseteq f^{-1}(A \cap B)$. The moral here is that preimages are well-behaved and images are poorly behaved. I'm not mathematically mature enough to articulate why that should be true, but it clearly is.

You can also check that 3 is always true directly. Again, preimages are well-behaved over most set operations. I can't think of an exception right now.

As you say, 4 is not always true. What is always true is that $$ f(f^{-1}(A)) \subseteq A $$ Again, try to think of an example that shows equality doesn't always hold. Hint: you can reuse $f(x) = x^2$ by choosing $A$ carefully. What property could $f$ have that would guarantee equality?

Finally, as an extraneous remark, it is also always true that $$ A \subseteq f^{-1}(f(A)) $$ At this point it might not surprise you that you can show equality won't always hold with $f(x) = x^2$ and a clever choice of $A$. What property could $f$ have that would guarantee equality?

Musings on Why Preimages Are Well-Behaved

We require that a function $f : A \to B$ is defined everywhere on its domain and is well-defined, in the sense that $a = b$ implies $f(a) = f(b)$. Definedness is the preimage version of surjectivity and well-definedness is the preimage version of injectivity. Injectivity and surjectivity suffice to make images behave well with respect to set operations. But all functions, by definition, already have the preimage version of injectivity and surjectivity. Perhaps this gives some accounting for why preimages are more well-behaved than images.

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    $\begingroup$ If you ever study topology, I recommend memorizing that $A \subseteq f^{-1}(f(A))$ and $f(f^{-1}(A)) \subseteq A$. $\endgroup$ – Charles Hudgins Sep 28 '19 at 4:01

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