Question

Let g be a function from set A to set B and f be a function from set B to set C. Assume that f °g is one-to-one and function f is one-to-one. Using proof by contradiction, prove that function g must also be one-to-one (in all cases).

Answer #1

Let f : A → B, g : B → C be such that g ◦ f is one-to-one (1 :
1).
(a) Prove that f must also be one-to-one (1 : 1).
(b) Consider the statement ‘g must also be one-to-one’. If it is
true, prove it. If it is not, give a counter example.

Let A be a finite set and f a function from A to A.
Prove That f is one-to-one if and only if f is onto.

Let A, B, C be sets and let f : A → B and g : f (A) → C be
one-to-one functions. Prove that their composition g ◦ f , defined
by g ◦ f (x) = g(f (x)), is also one-to-one.

1. Let A = {1,2,3,4} and let F be the set of all functions f
from A to A. Prove or disprove each of the following
statements.
(a)For all functions f, g, h∈F, if f◦g=f◦h then g=h.
(b)For all functions f, g, h∈F, iff◦g=f◦h and f is one-to-one
then g=h.
(c) For all functions f, g, h ∈ F , if g ◦ f = h ◦ f then g =
h.
(d) For all functions f, g, h ∈...

Let Let A = {a, e, g} and B = {c, d, e, f, g}. Let f : A → B and
g : B → A be defined as follows: f = {(a, c), (e, e), (g, d)} g =
{(c, a), (d, e), (e, e), (f, a), (g, g)}
(a) Consider the composed function g ◦ f.
(i) What is the domain of g ◦ f? What is its codomain?
(ii) Find the function g ◦ f. (Find...

Let f : A → B and suppose that there exists a function g : B → A
such that (g ◦ f)(a) = a and (f ◦ g)(b) = b. Prove that g = f −1
.
Thank you!

a) Let f : [a, b] −→ R and g : [a, b] −→ R be differentiable.
Then f and g differ by a constant if and only if f ' (x) = g ' (x)
for all x ∈ [a, b].
b) For c > 0, prove that the following equation does not have
two solutions. x3− 3x + c = 0, 0 < x < 1
c) Let f : [a, b] → R be a differentiable function...

Let A be a finite set and let f be a surjection from A to
itself. Show that f is an injection.
Use Theorem 1, 2 and corollary 1.
Theorem 1 : Let B be a finite set and let f be a function on B.
Then f has a right inverse. In other words, there is a function g:
A->B, where A=f[B], such that for each x in A, we have f(g(x)) =
x.
Theorem 2: A right inverse...

let F : R to R be a continuous function
a) prove that the set {x in R:, f(x)>4} is open
b) prove the set {f(x), 1<x<=5} is connected
c) give an example of a function F that {x in r, f(x)>4} is
disconnected

Let f:A→B and g:B→C be maps. Prove that if g◦f is a bijection,
then f is injective and g is surjective.*You may not use, without
proof, the result that if g◦f is surjective then g is surjective,
and if g◦f is injective then f is injective. In fact, doing so
would result in circular logic.

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