Given that f maps E to F and A is a subset of E, prove that if f is injective then f^-1(f(A)) is a subset of A.

Actually, im good at performing similar proofs, but i didn’t understand why should f be injective? I proved it without using this given and didn’t know how to benefit from it.

Please help me cz i have an exam.

Thanks ðŸ™‚

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If ff is not injective then there might be a,ba,b with aâˆˆAa\in A, bâˆ‰Ab\notin A and f(a)=f(b)f(a)=f(b).

â€“Â lulu

Oct 20 at 20:12

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But if f(x) Ã¢Ë†Ë† f(A) then for sure x Ã¢Ë†Ë† A. Can we consider f(x) Ã¢Ë†Ë† f(A) but x Ã¢Ë†â€° A ? Is that true even if f is not injective??

â€“Â Nour

Oct 20 at 20:26

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to prove that a set B is a subset of C, take an element of B and show it is in C.

â€“Â Abdallah Hammam

Oct 20 at 20:35

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No, f(x)âˆˆf(A)f(x)\in f(A) does not imply that xâˆˆAx\in A (unless ff is injective). Take f:Râ†’Rf:\mathbb R\to \mathbb R, f(x)=x2f(x)=x^2, A={2}A=\{2\}. Then f(A)={4}f(A) = \{4\}. But f(âˆ’2)=4f(-2)=4 so f(âˆ’2)âˆˆf(A)f(-2)\in f(A) even though âˆ’2âˆ‰A-2\notin A.

â€“Â lulu

Oct 20 at 21:34

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2 Answers

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Let xx be in fâˆ’1(f(A))f^{-1}(f(A)).

xâˆˆfâˆ’1(f(A))âŸ¹x\in f^{-1}(f(A)) \implies

f(x)âˆˆf(A)âŸ¹f(x) \in f(A) \implies

âˆƒaâˆˆA:f(x)=f(a)\exists a \in A : f(x)=f(a).

f(x)=f(a)f(x)=f(a) and ff injective âŸ¹ \implies

x=ax=a

x=ax=a and aâˆˆAâŸ¹xâˆˆAa\in A \implies x\in A

each xx of fâˆ’1(f(A))f^{-1}(f(A)) is in AA.

thus

fâˆ’1(f(A))âŠ‚Af^{-1}(f(A)) \subset A

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Why should I let a Ã¢Ë†Ë† A ? Why can’t I put x instead ?

â€“Â Nour

Oct 20 at 20:39

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f(A)f(A) contains, by definition, only images of the elements of AA.

â€“Â Abdallah Hammam

Oct 20 at 20:41

Choose xâˆˆfâˆ’1(f(A))x\in f^{-1}(f(A)). Then f(x)âˆˆf(A)f(x)\in f(A). Hence there exists aâˆˆAa\in A such that f(x)=f(a)f(x)=f(a). Injectivity of ff then implies that x=ax=a. In particular, xâˆˆAx\in A. Hence, fâˆ’1(f(A))âŠ‚Af^{-1}(f(A))\subset A.

To show that injectivity is really needed, consider for instance the non-injective function

f:{a,b,c}â†’{a,b,c}:{aâ†¦abâ†¦acâ†¦a.f:\{a,b,c\}\rightarrow\{a,b,c\}: \begin{cases} a\mapsto a\newline b\mapsto a\newline c\mapsto a \end{cases}.

If we take A={a}A=\{a\}, then fâˆ’1(f(A))=fâˆ’1({a})={a,b,c}f^{-1}(f(A))=f^{-1}(\{a\})=\{a,b,c\} which is not a subset of AA.

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Why should I let a Ã¢Ë†Ë† A ? Why can’t I put x instead ?

â€“Â Nour

Oct 20 at 20:38

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The equation f(x)âˆˆf(A)f(x)\in f(A) by itself does not necessarily imply that xâˆˆAx\in A. For instance, in the example I gave we have f(c)âˆˆf({a})f(c)\in f(\{a\}), though câˆ‰{a}c\notin \{a\}. All you can really say is that there exists some aâˆˆAa\in A such that f(x)=f(a)f(x)=f(a).

â€“Â studiosus

Oct 20 at 20:42

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Oh i got it. Thank you

â€“Â Nour

Oct 20 at 20:49