#Function composition
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It is about a characterization of injectivity.
but you asked why is fg = ..
what do you assume?
give the entire problem statement
I should prove that this statement is equivalent to the statement that f is injective, however, I don’t understand the statement. I thought that the input of a function is supposed to be the domain and the output the codomain. So that g(b)=a, and therefore, f(a)=b, which would be ∆B, but it’s not.
Yes
that statement is all sorts of whack then
We have a function f:A->B
f is injective if and only if there exists g: B->A such that gf = 1_A
What does 1_A mean?
Delta A
Ahh
1_A(a) = a for all a in A
For surjective it says g o f = 1_B
aL
do you define like this?
Yes, that is how i would define it
I think i need to ask my prof
f:A->B is surjective iff there exists g:B->A such that fg = 1_B
whenever you have composition gf = 1_A, then the first term is injective and last term is surjective
so f is injective since that's what you apply first
and g is surjective, becaus you apply it last
I think so too
this is nonsense already because B is not the domain of f
I guess I have to skip the homework and maybe reach out to my professor because I am absolutely confused how that can be
maybe he denotes compositions in reverse order in his class
regardless
this is not a complete problem statement
I have no clue
double check with classmates what the actual problem statement is
i refuse to believe that's all what your professor told you
Or maybe the denotation is really different
Thank you for your time, I appreciate your help. I’ll continue to look into it.
