Logical Structure of a Proposition

Question

I'm having a hard time figuring out the logical structure of the following theorem :

I'm not interested in proving it, for now, i'm just trying to understand its logical structure.
I don't know if it is a forward implication, perharps this one ( i never know which parts are not included in the statement, for example the parts of A,B,C are sets,A is emptyset, etc ) :

$\forall g,h ( f \circ g = f \circ h \to g = h ) \to f$ is one-to-one.

It seems so much like it is a definition for the one-to-one property of a function and i'm inclined to treat it like an equivalence although i know that i have not proven the converse yet.

It's similar to this one

Or this one :

Is this statement just like any other in the sense that its converse is radically different from its original version, or is there something particular of the logical structure of the theorem ?

I don't know how to describe exactly that logical structure , and i know it is a bit vague the converse, but in that particular case to me the converse doesnt seem radically different from the original version.This doesn't happen in the statement $" x$ is in paris $\to x$ is in france" or most other theorems in mathematics, where i find the converse a lot more different.

Can anyone give me an example of a statement with that particular logical structure ( that i tried to picture with the two examples ) such that the converse is false ?

Thanks in advance

Answer 1

Yes, the logical structure of the sentence is :

$\forall f \ [ \ \forall g \ \forall h \ (f∘g = f∘h \to g=h) \to \text {One-to-one}(f) \ ]$.

It is not the "usual" definition of Injective function but the bi-conditional holds, i.e. :

$f : B → C$ is injective if and only if, given any functions $g, h : A → B$, whenever $f∘g = f∘h$, then $g = h$.

If you want to "include" in the sentence also the sets $A,B,C$, you have to build a suitable conditional with the above sentence as consequent and the "prefix" involving them as antecedent, all being universally quantified.

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For the added example, the "formalization" must be :

$\forall f \ [ \ \exists g \ (g∘f = id_A) \to \text {One-to-one}(f) \ ]$.

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It is not clear to me your concern ...

You are absolutely right that from : $p \to q$ we are not licensed to infer : $q \to p$.

But (regarding your first example) if we do not use the bi-conditional as a definition of "one-to-one", in order to assert it we have to prove it.