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 and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
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

	$e = f^{- 1} \circ f = f^{- 1} \circ e \circ f = f^{- 1} \circ g^{- 1} \circ g \circ f =$ $= (f^{- 1} \circ g^{- 1}) \circ h = e$ $\Rightarrow h^{- 1} = {(g \circ f)}^{- 1} = f^{- 1} \circ g^{- 1}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum