**Proving a function f is one to one if and only if $f(Abigcap B) = f(A) bigcap f(B)$**without wasting too much if your time.

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Prove that if $f:Xrightarrow Y$ is a function then $f(Acap B) = f(A)cap f(B)$ for all subsets $A$ and $B$ if and only if $f$ is one to one.

Let $x,yin Abigcap B$ where $Abigcap B$ is not empty and $f(x) = f(y)$

$$xin Acap B rightarrow f(x)in f(Acap B) rightarrow f(x)in f(A)cap f(B)$$

$$yin Acap B rightarrow f(y)in f(Acap B) rightarrow f(y)in f(A)cap f(B)$$

Since $f(x) = f(y)$ then $f$ is one to one if and only if $f(Acap B) = f(A)cap f(B)$.

Not sure if this is right at all, if anyone can do a proof of this similar to what I have I would greatly appreciate it.

## Answer

$(Leftarrow)$Take $x,y in A$ such that $x neq y$, then ${x} cap {y} = emptyset$ then $emptyset =f(emptyset) = f({x}) cap f({y})$ then $f(x) neq f(y)$ and therefore $f$ is one-to-one.

**Proving a function f is one to one if and only if $f(Abigcap B) = f(A) bigcap f(B)$**- If you find the proper solution, please don't forgot to share this with your team members.