# (More) Incomplete problem statements concerning functions/compositions

First problem statement, as written:

Let $$f\colon B\rightarrow C$$ be a function. Prove that $$f$$ is injective if and only if, for every pair of functions $$g,h\colon A\rightarrow B$$, if $$f\circ g = f\circ h$$, then $$g=h$$.

This one seems easy to spot. Clearly $$A$$ must be non-empty, as otherwise a contradiction may easily be reached (not to mention the proof is impossible otherwise).

[If $$A$$ is empty, then $$g,h$$ are empty. If we’re working in the direction to “prove” $$f$$ is injective, then (vacuously) if $$f\circ g = f\circ h$$ then $$g=h$$. But if $$f$$ is any constant function with a domain of at least two elements (which remains possible because even though $$A$$ being empty forces $$g,h$$ to be empty, their codomain, $$B$$, need not be empty to satisfy the definition of function), then f is not injective, even though all premises hold.] No?

Second problem statement, as written:

Let $$f\colon A\rightarrow B$$ be a function. Prove that $$f$$ is surjective if and only if, for every pair of functions $$g,h\colon B\rightarrow C$$, if $$g\circ f = h\circ f$$, then $$g=h$$.

For this problem, there seem to be two issues on restrictions on the sets the functions act on: 1) either $$A,B$$ are both empty, or both non-empty, and 2) $$C$$ must contain at least two elements.

I concluded 1) since, if $$B$$ is empty, $$A$$ must be empty to satisfy the definition of function. If $$A$$ alone is empty, then $$f$$ cannot be surjective (either by hypothesis, or as a conclusion).

I concluded 2) since I got to a junction in my proof in which I was unable to proceed without additional premises. After considering some possibilities, the contention that $$C$$ contain at least two elements seemed a plausible way out of this. [for the record, in “proving” $$f$$ to be surjective, I supposed the antecedent if the conditional (in that direction). Next, I proceeded by contradiction, supposing (to the contrary) $$g\ne h$$. If $$C$$ is unspecified (but non-empty, as otherwise $$A,B,C$$ are all forced to be empty, and the theorem becomes vacuous), it seems impossible to define $$g,h$$ by which to arrive at a contradiction.] Am I correct?

• Needs more focus: This question currently includes multiple questions in one. It should focus on one problem only. Commented May 6, 2020 at 3:12

The only if direction says, this has to hold for every such pair of functions, not just one example. So you just showed that sure, for $$A$$ being the empty set, this gives you no extra information about $$f$$. But let's assume like you said that $$f$$ is a constant function with a domain with e.g. two elements, so $$f : \lbrace b_1, b_2 \rbrace \to \lbrace c_1, c_2 \rbrace$$ and $$f(b_1) = f(b_2) = c_1$$. Now let's test the assumption with the two functions $$g,h : \lbrace a \rbrace \to \lbrace b_1, b_2 \rbrace$$ defined as $$g(a) = b_1$$ and $$h(a) = b_2$$, so clearly $$g \neq h$$. But $$f \circ g(a) = f(b_1) = c_1$$ and $$f \circ h(a) = f(b_2) = c_1$$, which means $$f \circ g = f \circ h$$, so these two "test functions" show that $$f$$ is not injective.