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Knowing that $f: X \to Y \text{ and } g:Y \to Z$ are $1-1$ and surjective,are then also $f^{-1}: Y \to X$ and $g \circ f:X \to Z$ $1-1$ and surjective or just $1-1$ ?

evinda
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1 Answers1

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1) $f^{-1}$ is surjective and 1-1, because for every $x \in X \ \ f^{-1}(f(x)) = x$ and if $f^{-1}(a) = f^{-1}(b)$ then $a = f(f^{-1}(a)) = f(f^{-1}(b)) = b$

2) $g \circ f $ is surjective and $1-1$, because if $g \circ f (a) = g \circ f (b) $ then $f(a) = f(b) $ and so $a = b$. Moreover for every $z \in Z $, let $b \in Y $ be such that $g(b) = z $ and $a \in X $ such that $f(a) = b $. Then $g \circ f (a) = z$

WLOG
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    Bijective means $1-1$ and surjective, so your first sentence is redundant. I assume you meant "surjective and $1-1$." – Thomas Andrews May 01 '14 at 14:40
  • To go from the relation $ f^{-1} (a)= f^{-1} (b) $ to the relation $a=f (f^{-1}(a))= f (f^{-1}(b))=b$ don't we have to know that $f^{-1}$ is invertible? – evinda May 01 '14 at 14:47
  • yes, infact $(f^{-1})^{-1} = f$ – WLOG May 01 '14 at 14:56