I'm about 70 pages into Lawvere & Schanuel's book Conceptual Mathematics and came across a question while solving an exercise, which at first seemed trivial but after toying around with it some more left me rather confused. I shall rephrase and expand upon the question as follows:
Throughout, let $\mathscr{C}$ be a category with a terminal object $\textbf{1}$.
Let $X\stackrel{\tilde{f}}{\longrightarrow}Y$ be a morphism in $\mathscr{C}$ such that $\tilde{f}\circ g=f$ for some morphisms $Z\stackrel{f}{\longrightarrow}Y$ and $Z\stackrel{g}{\longrightarrow}X$, i.e. $\tilde{f}$ is an extension of $f$ along $g$. It follows immediately that for any pair $\textbf{1}\stackrel{z_1, z_2}{\rightrightarrows}Z$ of global elements, one has the implication:
$$g(z_1)=g(z_2) \implies f(z_1)=f(z_2)$$
My question being, is the converse statement true? And if not, under what additional assumptions will it be true? In other words, if two morphisms $f$ and $g$ satisfy the above implication, must there necessarily exist an extension of $f$ along $g$? (Naturally every monic $g$ will trivially satisfy the implication.)
I figured I would simply construct a counterexample and disprove the claim, however all of the examples I managed to conjure up so far have only hinted towards the affirmative. Though I suspect there is something quite straightforward I ought to have overlooked.
An analogous question arises when examining the dual notion of a $\textit{lift}$, where the situation is practically the same as above. I will not describe it here, as I suspect the answer to the former will reveal the answer to the latter. Any hints are appreciated.
EDIT: I apologize for the confusion. I did not mean to imply that I was seeking the converse implication, but rather I am assuming the original implication and seeking to construct an extension.
Regarding the converse implication:
Given $f:Z\rightarrow Y, g:Z\rightarrow X$ and $\widetilde{f}:X \rightarrow Y$ with $f=\widetilde{f}g$ and global points $z_1,z_2:1\rightarrow Z$ then clearly $$gz_1 = gz_2 \Rightarrow fz_1 = \widetilde{f}gz_1 = \widetilde{f}gz_2 = fz_2$$ The converse does not hold in general, for example in the case $\mathscr{C}=\mathsf{Set}$ with $Z=X=\{0,1\}, Y=\{*\},f,\widetilde{f}$ constant and $g=\operatorname{id}$. However if $\widetilde{f}$ is monic we obtain $$fz_1 = fz_2 \Leftrightarrow \widetilde{f}gz_1 = \widetilde{f}gz_2 \Rightarrow gz_1 = gz_2$$
Regarding the existence of an extension:
Take $\mathscr{C}=\mathsf{Top}$ with $Z=Y=\{0,1\}_\text{disc}, X =\{0,1\}_\text{chaot}$ and $f,g=\operatorname{id}$ on underlying sets. Then obviously the implication holds, but there cannot be an extension $\widetilde{f}:X=\{0,1\}_\text{chaot} \rightarrow Y=\{0,1\}_\text{disc}$ by continuity reasons.