Let $g\circ f = h\circ f$ for all $f$ from $A$ to $B$, prove that $g = h$

Question

Let $g:B \to C$ and $h: B \to C$ be functions, so that $g \circ f = h\circ f$ for all $f: A \to B$. To prove that $g=h$. The way I proceeded to prove is as follows:

Since $g\circ f = h\circ f$ for all $f$, I thought it was safe to assume that $f$ is surjective, so that $B \subset ranf$. Hence:

$(x,y) \in g \implies x \in domg \implies x \in B \implies x\in ranf \implies \exists z(z,x) \in f \implies \exists z (z,x) \in f $and $(x,y) \in g \implies \exists z(z,y) \in g\circ f \implies \exists z(z,y) \in h\circ f \implies \exists v, \exists z (z,v) \in f $and $(v,y) \in h \implies \exists z, \exists v\in f $ and $(z,x) \in f \implies v=x \implies (x,y) \in h$.

The second half of the proof is completely analogous to the first half, so I don't think it's required to write it down. However, did I prove it correctly? Was there any flaw or falt in my reasoning? I am asking for I require constant confirmation from the more professional since I am only new to set theory. Thank you in advance.

Answer 1

Your proof cannot work, because you assume that there exists $f : A\to B$ that is surjective. Counter-example: there is no surjection from $\mathbb{N}$ to $\mathbb{R}$.

Second point, there is something missing in your statement : the initial quantifier. Indeed, the goal is to prove $$\forall x\in B\,,g(x)=h(x)$$ or, as you wrote using "formal ZF function definition", $$\forall (x,y)\in B\times C\,,(x,y)\in g\Leftrightarrow (x,y)\in h$$

You must begin with this "forall" quantifier.

However, you can process as this (supposing neither $A,B,C$ are empty) :

• $\forall x\in B$, let $f_x : A\to B$ be the constant function : $\forall z\in A\,,f_x (z)=x$.
• Then you'll come to $g\circ f_x(z)=h\circ f_x(z)$
• and finally $g(x)=h(x)$,
• thus $g=h$ because at the begining there was $\forall x\in B$

Answer 2

I want to pull out some things from the comments under the question. As peterwhy noted, it's not clear if $ A $ is given ahead of time; that is, are you meant to prove $$ \forall \, B , \; \forall \, C , \; \forall \, g \colon B \to C , \; \forall \, h \colon B \to C , \; \forall \, A , \; ( \forall \, f \colon A \to B , \; g \circ f = h \circ f ) \; \Rightarrow \; g = h $$ or $$ \forall \, B , \; \forall \, C , \; \forall \, g \colon B \to C , \; \forall \, h \colon B \to C , \; ( \forall \, A , \; \forall \, f \colon A \to B , \; g \circ f = h \circ f ) \; \Rightarrow \; g = h \text ? $$ The difference is that (in proving the statement) you get to pick $ A $ in the second case but not in the first.

The second one is very easy to prove; your proof works, but so does the simpler one in a comment by Thomas Andrews: let $ A $ be $ B $, and let $ f $ be the identity function. So that's probably not the intended interpretation; it's too easy.

The first one is false, for a reason first pointed out implicitly in a comment by Severin Schraven: let $ B $ be any set with at least one element, let $ C $ be any set with at least two elements, let $ g $ and $ h $ be different functions from $ B $ to $ C $ (such as the constant functions to different elements of $ C $), and let $ A $ be empty. So that can't be it; too hard (impossible).

But many of the comments, including the one by Severin Schraven (which contains a complete answer), and also the answer by hdci, assume a modified version of the first interpretation, where we additionally assume that $ A $ is inhabited (meaning non-empty). So that's $$ \forall \, B , \; \forall \, C , \; \forall \, g \colon B \to C , \; \forall \, h \colon B \to C , \; \forall \, A , \; A \ne \varnothing \; \Rightarrow \; ( \forall \, f \colon A \to B , \; g \circ f = h \circ f ) \; \Rightarrow \; g = h \text . $$ This is the interpretation that's both true and not too easy; if this question came from an assignment, then you might want to check whether you were supposed to assume that $ A $ is inhabited. (It's not necessary to assume that $ B $ and $ C $ are inhabited.)

This isn't a full answer, since other people gave the proofs for the true interpretations, but I wanted to make it clear that the nontrivial interpretation isn't quite true, and not just leave that to parentheses.