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Question Number 109149 by mohammad17 last updated on 21/Aug/20
if f:x→x be a mapping prove that (f⊆I_X ∨I_X ⊆f)→f=I_(X ) ?  help me sir
$${if}\:{f}:{x}\rightarrow{x}\:{be}\:{a}\:{mapping}\:{prove}\:{that}\:\left({f}\subseteq{I}_{{X}} \vee{I}_{{X}} \subseteq{f}\right)\rightarrow{f}={I}_{{X}\:} ? \\ $$$${help}\:{me}\:{sir} \\ $$
Commented by kaivan.ahmadi last updated on 21/Aug/20
X→^f X  X→^I X , I(x)=x  if   f⊆I_X  then for each x in X we haveI_X (x)=x  on the other hand x=f(y) for some y in X  so I_X ⊆f and I_X =f.  if I_X ⊆f then for each x in X we have f(x)∈X  let f(x)=y∈X and so f(x)=y=I_X (y) so f⊆I_X , hence  f=I_X .
$${X}\overset{{f}} {\rightarrow}{X} \\ $$$${X}\overset{{I}} {\rightarrow}{X}\:,\:{I}\left({x}\right)={x} \\ $$$${if}\:\:\:{f}\subseteq{I}_{{X}} \:{then}\:{for}\:{each}\:{x}\:{in}\:{X}\:{we}\:{haveI}_{{X}} \left({x}\right)={x} \\ $$$${on}\:{the}\:{other}\:{hand}\:{x}={f}\left({y}\right)\:{for}\:{some}\:{y}\:{in}\:{X} \\ $$$${so}\:{I}_{{X}} \subseteq{f}\:{and}\:{I}_{{X}} ={f}. \\ $$$${if}\:{I}_{{X}} \subseteq{f}\:{then}\:{for}\:{each}\:{x}\:{in}\:{X}\:{we}\:{have}\:{f}\left({x}\right)\in{X} \\ $$$${let}\:{f}\left({x}\right)={y}\in{X}\:{and}\:{so}\:{f}\left({x}\right)={y}={I}_{{X}} \left({y}\right)\:{so}\:{f}\subseteq{I}_{{X}} ,\:{hence} \\ $$$${f}={I}_{{X}} . \\ $$$$ \\ $$
Commented by mohammad17 last updated on 21/Aug/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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