disprove of statements

[sharp vault]

For the first statement

A = C = {a}, B = {b, c}, let be f(a) = b, g(b) = g(c) = a

I mean both statements are wrong and I have to disprove it. Does this work ? And how can I explain it in a right way that it works ?

[crude cypress]

if gof injective then f injective

and if gof surjective then g surjective

those are true

but yeah you clearly understood that the statements you were given are false

to disprove them you just have to find a counter-example

[sharp vault]

I tried it

But it seems there is smth wrong

[crude cypress]

I am trying to work through both of your attempts seeing if it works or not

[sharp vault]

g:R -> [0,1] isn’t defined for every point ig…..

[sharp vault]

Ayo thx 😭

[thin rock]

@earnest temple has given 1 rep to @crude cypress

[crude cypress]

I mean you seem to have gotten it right there

gof : {a} -> {a} gof(a) = a is clearly bijective

so gof is injective but g is not injective since g(b) = g(c)

and gof is also surjective but f is not since there is no value associated with c

[crude cypress]

that example is also perfect

[0, pi/2] -> [-1,1] -> [0,1]

sin^2 is bijective from [0, pi/2] to [0,1]

but (-1)^2 = 1^2 so g(x) = x^2 is not injective

and -1 is not in sin([0,pi/2]) so f(x) = sin(x) isn't surkective from [0,pi/2] to [-1,1]

@sharp vault Your examples work like wonders but what you need to say is that f is NOT surjective and g is NOT injective to disprove the statement

[sharp vault]

I got it

And I did this for the second statement

[crude cypress]

@sharp vault you didn't need to

you first (or second examples) work for both statements

[sharp vault]

Because my first example work for both ?

[sharp vault]

Oh yeahh

[sharp vault]

Second too ?