Let g : A -> B and f : B -> C be functions.
Show that if f and g are injective, then f o g is injective.
Show that if f and g are surjective, then f o g is surjective.
since f is surjective, rng f = C. That is, for any c C, there exists b B such that f(b) = c. Now since g is surjective, there exists an a A such that g(a) = b. but then (f o g)(a) = f(g(a)) = f(b) = c, so
f o g is surjective.
Let g : A -> B and f : B -> C be functions.
Show that if f and g are injective, then f o g is injective.
Show that if f and g are surjective, then f o g is surjective.
i don't know how to proove it
For to be injective . Assume . Since is injective...
Spoiler:
Since is injective, . Since is injective, , so , and is injective.
For to be surjective, , such that . Let . Because is surjective...
Spoiler:
Because is surjective, there exists such that . Because is surjective, there exists such that . So and is surjective.