Question

4.2. Let $f: X \to Y$ be a function from a set X to a set Y. Show that (1) f is injective if and only if f has a left inverse $g: Y \to X$ such that $g \circ f = id_X$, the identity function on X, (2) f is surjective if and only if f has a right inverse $g: Y \to X$ such that $f \circ g = id_Y$, the identity function on Y, (3) f is bijective if and only if f has a two-sided inverse $g: Y \to X$ such that $g \circ f = id_X$ and $f \circ g = id_Y$.

          4.2. Let $f: X \to Y$ be a function from a set X to a set Y. Show that
(1) f is injective if and only if f has a left inverse $g: Y \to X$ such that $g \circ f = id_X$, the identity function on X,
(2) f is surjective if and only if f has a right inverse $g: Y \to X$ such that $f \circ g = id_Y$, the identity function on Y,
(3) f is bijective if and only if f has a two-sided inverse $g: Y \to X$ such that $g \circ f = id_X$ and $f \circ g = id_Y$.

4.2. Let f: X → Y be a function from a set X to a set Y. Show that
(1) f is injective if and only if f has a left inverse g: Y → X such that g ∘ f = idX, the identity function on X,
(2) f is surjective if and only if f has a right inverse g: Y → X such that f ∘ g = idY, the identity function on Y,
(3) f is bijective if and only if f has a two-sided inverse g: Y → X such that g ∘ f = idX and f ∘ g = idY.

Added by Joseph R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

07/26/2023

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Step 1: We need to prove that a function f is injective if and only if it has a left inverse. Show more…

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(5) $f(x) = |x|$. 4.2. Let $f: X \rightarrow Y$ be a function from a set X to a set Y. Show that (1) f is injective if and only if f has a left inverse $g: Y \rightarrow X$ such that $g \circ f =$ $id_X$, the identity function on X, (2) f is surjective if and only if f has a right inverse $g: Y \rightarrow X$ such that $f \circ g = id_Y$, the identity function on Y, (3) f is bijective if and only if f has a two-sided inverse $g: Y \rightarrow X$ such that $g \circ f = id_X$ and $f \circ g = id_Y$.