If f is injective can you prove this relation?

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 🙂

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If ff is not injective then there might be a,ba,b with a∈Aa\in A, b∉Ab\notin A and f(a)=f(b)f(a)=f(b).
– lulu
Oct 20 at 20:12

  

 

But if f(x) ∈ f(A) then for sure x ∈ A. Can we consider f(x) ∈ f(A) but x ∉ A ? Is that true even if f is not injective??
– Nour
Oct 20 at 20:26

  

 

to prove that a set B is a subset of C, take an element of B and show it is in C.
– Abdallah Hammam
Oct 20 at 20:35

  

 

No, f(x)∈f(A)f(x)\in f(A) does not imply that x∈Ax\in A (unless ff is injective). Take f:R→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(−2)=4f(-2)=4 so f(−2)∈f(A)f(-2)\in f(A) even though −2∉A-2\notin A.
– lulu
Oct 20 at 21:34

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2 Answers
2

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Let xx be in f−1(f(A))f^{-1}(f(A)).

x∈f−1(f(A))⟹x\in f^{-1}(f(A)) \implies

f(x)∈f(A)⟹f(x) \in f(A) \implies

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

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

x=ax=a

x=ax=a and a∈A⟹x∈Aa\in A \implies x\in A

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

thus

f−1(f(A))⊂Af^{-1}(f(A)) \subset A

  

 

Why should I let a ∈ A ? Why can’t I put x instead ?
– Nour
Oct 20 at 20:39

  

 

f(A)f(A) contains, by definition, only images of the elements of AA.
– Abdallah Hammam
Oct 20 at 20:41

Choose x∈f−1(f(A))x\in f^{-1}(f(A)). Then f(x)∈f(A)f(x)\in f(A). Hence there exists a∈Aa\in A such that f(x)=f(a)f(x)=f(a). Injectivity of ff then implies that x=ax=a. In particular, x∈Ax\in A. Hence, f−1(f(A))⊂Af^{-1}(f(A))\subset A.

To show that injectivity is really needed, consider for instance the non-injective function
f:{a,b,c}→{a,b,c}:{a↦ab↦ac↦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 f−1(f(A))=f−1({a})={a,b,c}f^{-1}(f(A))=f^{-1}(\{a\})=\{a,b,c\} which is not a subset of AA.

  

 

Why should I let a ∈ A ? Why can’t I put x instead ?
– Nour
Oct 20 at 20:38

  

 

The equation f(x)∈f(A)f(x)\in f(A) by itself does not necessarily imply that x∈Ax\in A. For instance, in the example I gave we have f(c)∈f({a})f(c)\in f(\{a\}), though c∉{a}c\notin \{a\}. All you can really say is that there exists some a∈Aa\in A such that f(x)=f(a)f(x)=f(a).
– studiosus
Oct 20 at 20:42

  

 

Oh i got it. Thank you
– Nour
Oct 20 at 20:49