Help with a proof on injectivity

[past loom]

I had my exam on logic and proofs today and im just wondering if this was a valid starting approach

[past loom]

The rest of the proof i dont need help with pretty certain i had done it right

[valid pasture]

Huh?

where's the approach

Also is that meant to be ∀A ⊂ X

and B ⊂ Y

[past loom]

Yeah my bad

Basically idk if im allowed to write the equivalence to an implication like that

I dont see why not tho

Because if i assume the left side to then use properties of injectivity to get a logical implication to the right

And do the same for the other way around

That works no?

Ill write the entire thing down

[valid pasture]

well

it is technically a true statement but I don't see how it can help here

[past loom]

Well thats good atleast

Its because i just didnt know how to properly prove it from A = f-1(B) itself it that makes sense?

But im writing it down

[valid pasture]

I think the easiest route is probably this

assume f injective then prove the LHS holds

then assume f not injective and prove the LHS fails

[past loom]

There u go

Probably still some mistakes

I wrote it down pretty fast

[valid pasture]

nice handwriting

the first part (⇒) is partially correct

[past loom]

Thank u

[valid pasture]

the second part (⇐) has some logical errors and it is not too clear what you are proving

Since you want to prove the LHS result, you should start by letting A ⊂ X be arbitrary, and then construct a B so that f^-1(B) = A.

[past loom]

I knew i went wrong on that part where i picked B

My intuition was that for whatever B i picked i couldve made an A for which it ends up with the same issue

And i think i wrote that down too on the exam

But otherwise id have no clue

[valid pasture]

So you want to show that, for this choice of A, there exists no B so that f^-1B = A

you can analyse two possibilities: B contains f(x1), and B does not contain f(x1)

[past loom]

And if it doesnt contain f(x1) it goes against the assumption that A is subset of f-1(B)?

[valid pasture]

if B doesn't contain f(x1), then x1 cannot possibly be in f^-1(B), therefore f^-1(B) cannot be equal to A = {x1}

[past loom]

Makes sense

[valid pasture]

try the other case now, where B contains f(x1)

[past loom]

Yeah so youre left with the fact that f-1(B) would always contain an element thats not in A because of the non injectivity

[valid pasture]

yes, to be precise, if B contains f(x1), then f^-1B must contain both x1 and x2, since f(x1)=f(x2), and therefore f^-1 cannot possibly be the singleton {x1}

[past loom]

Its sad because i was going to do those two cases but for some reason i couldnt figure out what to do when B didnt contain f(x1)

And now it sounds so simple too

As long as i dont get a 0 for that part tbh

[valid pasture]

No worries, it is very common in maths that before you understand something it seems impenetrably difficult, but after you understand it everything suddenly becomes clear and trivial