Question

Find functions f,g:N-->N, so that the following are true at the same time:
1. f is not surjective
2. g is not injective
3. g o f is bijective

I have been trying to figure this out for hours and nothing seems to work!!

Answer 1

Not sure, but how about these functions?

\(f(x) = 2+\sqrt {x-1}\\~\\g(x) = (x-2)^2+1\)

Answer 2

Maybe somehow have function be piecewise-defined?

Answer 3

g(x) isn't surjective either though (eg you can't get g(x)=3).. so how is gof gonna be bijective?

Answer 4

Well question never said that \(g\) has to be surjective.

Also \((g\circ f)(x) = x\) for \(x\ge1\), which makes it bijective, right?

Answer 5

well g(x) necessarily has to be surjective for gof to be surjective....

Answer 6

I mean definition says that for gof to be bijective f has to be injective and g surjective

Answer 7

Well I'm out of luck. Maybe these users can help: @amistre64 @TuringTest @perl

Or you can ask this question on math stack exchange?

Answer 8

Will give that a go, thanks!

Answer 9

bijective means 1-1 and onto

surjective in onto, injective is 1-1

just trying to remember the defs

Answer 10

f is not surjective; or f is injective (1-1 but not onto)

f(n) = 2n is clearly 1-1 but not onto

Answer 11

surjective is onto, but need not be 1-1

simplest thing is to right shift a parabola

g(n) = (n-1)^2 , does n include 0? if not up it by 1

Answer 12

oh, you are trying to find f and g such that all 3 conditions hold at the same time

Answer 13

the question was to find g(f(x)) which is one to one correspondence
regardless of f and g i guess

Answer 14

g(f) = n + k is superb for bijective lol

Answer 15

well, k=0 maybe

Answer 16

g and f should be inverses then, or at least one way inverses