Given-f-E-F-g-F-G-E-F-and-G-are-sets-Show-that-if-f-and-g-are-bijectives-then-g-f-is-bijective-and-g-f-1-f-1-g-1-

Question Number 165576 by mathocean1 last updated on 04/Feb/22

G i v e n f : E \to F; g : F \to G .

E; F a n d G a r e s e t s .

S h o w t h a t i f f a n d g a r e b i j e c t i v e s

t h e n g \circ f i s b i j e c t i v e a n d

{(g \circ f)}^{- 1} = f^{- 1} \circ g^{- 1}

Answered by aleks041103 last updated on 04/Feb/22

f i s b i j . \Leftrightarrow f i s i n j . a n d s u r j .

l e t h = g \circ f : E \to G

\underset{―}{1 . h i s s u r j .}

h (E) = g (f (E))

f - s u r j . \Rightarrow f (E) = F

h (E) = g (F)

g - s u r j . \Rightarrow g (F) = G

\Rightarrow h (E) = G

\Rightarrow h i s s u r j .

\underset{―}{2 . h i s i n j .}

t a k e a n y x \in G .

g i s b i j . \Rightarrow! \exists y \in F, g (y) = x

f i s b i j . \Rightarrow! \exists z \in E, f (z) = y

\Rightarrow x = g (y) = g (f (z)) = h (z)

\Rightarrow \forall x \in G,! \exists z \in E, h (z) = x

\Rightarrow h i s i n j .

\Rightarrow \underset{―}{h i s b i j .}

Answered by aleks041103 last updated on 04/Feb/22

e = f^{- 1} \circ f = f^{- 1} \circ e \circ f = f^{- 1} \circ g^{- 1} \circ g \circ f =

= (f^{- 1} \circ g^{- 1}) \circ h = e

\Rightarrow h^{- 1} = {(g \circ f)}^{- 1} = f^{- 1} \circ g^{- 1}