Let A,B, and C be nonempty sets. Let f: A→B and g: B→C be functions. For eac

Question

Let A,B, and C be nonempty sets. Let f: A→B and g: B→C be functions. For each of the following statements, either provide a PROOF or give a specific (simple) COUNTEREXAMPLE.<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

a) f is injective→ g o f is injective

b) g is injective→ g o f is injective

c) f is surjective→ g o f is surjective

d) g is surjective→ g o f is insurjective

e) g o f is injective→ f is injective

f) g o f is injective→ g is injective

g) g o f is surjective→ f is surjective

h) g o f is surjective→ g is surjective

Explanation / Answer

(a) counter example

A = {0,1} B = {0,1} C = {0,1}

f(x) = x for x in A , g(x) = 0 for x in B

Then gof : A -> C is defined by gof(x) = 0 for x in A

Thus f is injective but gof is not injective.

(b) Counter example

A = {0,1} B = {0,1} C = {0,1}

f(x) = 0 for x in A , g(x) = x for x in B

then gof(x) = 0 for x in A

Thus g is injective but gof is not.

(c) counter example

A = {0,1} B = {0,1} C = {0,1}

f(x) = x for x in A , g(x) = 0 for x in B

Then gof : A -> C is defined by gof(x) = 0 for x in A.

f is surjective but gof is not

(d) counter example

A = {0,1} B = {0,1} C = {0,1}

f(x) = 0 for x in A , g(x) = x for x in B

then gof(x) = 0 for x in A

Thus g is surjective but gof is not surjective

(e) True,

Let x,y be two distinct elements in A.Since gof is injective gof(x) and gof(y) are distinct.

i.e g[f(x)] is not equal to g[f(y)] => f(x) is not equal to f(y).Hence f is injective.

(f) Counter example

A = {0}, B = {0,1} , C = {0,1}

f(0) = 0 , g(x) = 0 for x in B

Then gof is injective (trivially since A has only one element)

but g is not injective g(0) = 0 = g(1)

(g) Counter example

A = {0} , B = {0,1} , C = {0}

f(0) = 0 , g(x) = 0 for x in B

Then gof is surjective gof(0) = 0 and 0 is the only element in C

But f is not surjective.There is no preimage for 1

(h) True

Let c be an element in C.

since gof is surjective there exists a in a such that gof(a) = c

=> g[ f(a) ] = c

=> g ( b ) = c ,where b = f(a) belongs to B

Hence given c in C there exists b in B such that g(b) = c.

Hence g is surjective.

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