For unlimited access to Class Notes, a Class+ subscription is required.

November 13 2014

ETLC 2-001

f B is a function a!b f

Function f: A -> B is 1-1 a1,a2 a1 = a2

is onto ba C

gaf(a) = g(f(a))

Example:

A={1,2,3,4}

B={a,b,c}

C={a,b,c,d}

D={5,6,7,8,9}

f={(1,a),(2,a),(3,a),(4,b)}

g={(a,9),(b,8),(c,7),(d,5)}

gg = undefined

Just wrong

f={(1,c),(2,c),(3,c),(4,b)}

g={(a,9),(b,8),(c,7),(d,5)}

gg = undefined

Example:

a:

a: (x,y) -> x + y

r4: a: a: -> r4(a(x,y))

Theoroms: For functions f: A -> B and g: C -> D where B f is

also 1-1

Suppose f and g are 1-1 and ga1,a2 a1 = a2

Let a1, a2 f(a1) = g A, f(a1) = f(a2) f is also onto.

A, f(a1) = f(a2) f is also onto.

Suppose B = C and f and g are onto.

Suppose B = C and f and g are onto.