I need some help..

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let's first deal with one-to-one: let \(S_1,S_2\in D\) such that \[ \large f(S_1)=f(S_2) \] we have to prove that \(S_1=S_2\). we have four cases: \(\mathbf{(1)}\) \(1\in S_1\qquad 1\in S_2\) then \[ \large f(S_1)=f(S_2) \] \[ \large S_1\setminus\{1\}=S_2\setminus\{1\} \] \[ \large S_1=S_2 \] \(\mathbf{(2)}\) \(1\notin S_1\qquad 1\notin S_2\) then \[ \large f(S_1)=f(S_2) \] \[ \large S_1\cup\{1\}=S_2\cup\{1\} \] \[ \large S_1=S_2 \]
Do we need to prove if \[ S_1 \not=S_2\quad then \quad f(S_1)\not=f(S_2)\] for one to one
actually we can use both of them..
these are definition.. \[ \forall a,b \in A, \;\; f(a)=f(b) \Rightarrow a=b \\ \forall a,b \in A, \;\; a \neq b \Rightarrow f(a) \neq f(b) \]
this is definition for onto \[ \forall y \in Y, \, \exists x \in X, \;\; f(x)=y \]
thanks. but i know the definitions.
onto is easier: let \(K\in E\) if \(1\notin K\) then \[ \large K=f(K\cup\{1\}) \] if \(1\in K\) then \[ \large K=f(K\setminus\{1\}) \]
thanks a lot..
what are the other two cases..

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