Can functoriality be explained in the opposite way?

Question

I have some confusion about functoriality in category theory.

In general, functoriality means that functors must preserve composition of morphisms.

That means

Given a functor $F :C \to D$, for every morphisms $f:A \to B$ and $g:B \to C$ in C, we have $F(g) \circ F(f) = F(g \circ f)$.

I have two questions for this axiom.

The first question is

Given a functor $F :C \to D$, if we know that $\exists$ morphisms $f, f', f''$ in $C$ make $F(f) \circ F(f') = F(f'')$ in D, can we conclude that $f \circ f' = f''$ in C? (It means that we can elimiate $F$ at the both side)

The second question is

If the answer is NO, what conditions f f’ f’’ must satisfy to make the above conclusions YES?

Thanks.

Answer 1

In general, if all you know is that $F(f)\circ F(f')=F(f'')$, the composition $f\circ f'$ may not even be defined, and even if it exists it need not coincide with $f''$. For example, if $X,Y,Z$ are objects of $C$ for which $F(X)=F(Y)=F(Z)$, then $F(id_X)\circ F(id_Y)=F(id_Z)=F(id_X)\circ F(id_X)$, but $id_X\circ id_Z$ is not defined, and $id_X\circ id_X$ is defined but distinct from $id_Z$. The only way it would work is if $F$ is injective on arrows (and thus also on objects).

If you impose that $f,f',f''$ are arrows $X\to Y$, $Y\to Z$ and $X\to Z$ respectively, then your question is equivalent to the following one :

If $F: C\to D$ is a functor and if $f,g:X\to Y$ are morphisms in $C$ such that $F(f)=F(g)$, is it true that $f=g$ ?

In general, the answer is no, there is no reason to assume this. For example, monoid homomorphisms are not always injective. A functor for which the answer is "yes" is called a faithful functor.