functions

Question

Let $f:A\to B$ and $E$ and $F$ be subsets of $A$. Is below statement true or false?

$S:f(E \cap F)= f(E) \cap f(F)$

Comments

Similar type of question asked in Gate_2001_2.3

Answer 1

False :

Consider $E: \{1,2,3,4\}$,   $F:\{3,4,5,6\}$

$f(E) = \{5,6,7,8\}$

$f(F)=\{7,8,11,6\}$

$f(E \cap F) = \{7,8\}$

where as, $f(E) \cap f(F) = \{6,7,8\}$

Becomes TRUE only when $f$ is one-one (injective). 

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Alternatively

$ f(E \cap F) \subseteq f(E)$ as intersection never adds any new element to a set and a function should have a mapping for all elements in domain.

Similarly, $ f(E \cap F) \subseteq f(F)$

Combining both,

$ f(E \cap F) \subseteq f(E) \cap f(F)$

$\subseteq$ becomes $=$ only when $f$ is one-one - when the mappings are unique for each element. 

Comments

I can t understand it as E F are domain and codomain and f(E) f(F) are images of these so why elements of E F and f(E) and f(F) are different

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because some elements of E and F can be mapped to the same element. You can see the given example.

Answer 2

Yes if f is one-one and onto else no

Comments

Can you please explain with example....??

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Why is onto required?

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@ Arjun Sir, Can you elaborate above proof? how can we decide nature of function f:A->B

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I don't have a proof. To be true the function must be one-one. Otherwise it is easy to produce a counter example as shown in answer.