Composition of arrows on slice category

Question

As the definition says, the arrow $g$ is an arrow from $X$ to $X'$ such that $f'\circ g = f$. I'm trying to imagine an arrow $g:X\to X'$ such that the composition with $f'$ does not give $f$. Is it possible?

Answer 1

Let's do it in category $\mathbf{Set}$.

Then you need functions $f:X\to C$, $f':X'\to C$ and $g:X\to X'$ such that: $$f'\circ g\neq f$$ You could take $X$ as a singleton and $f$ as the function that sends its unique element to some $c\in C$.

Further let $f':X'\to C$ be a function with $c\notin\mathsf{im}f'$.

Answer 2

Take C to be the free category generated by the graph in your picture. It is the category with objects $X, X', C$, and whose arrows are paths in the diagram. Composition is given by concatenating paths, and the identity arrows are paths of length zero.

$\hom(X, C)$ has exactly two arrows. One is the path $f$, and the other is the path $f'g$. (i.e. $f'g$ is the path that starts from $X$, follows the edge $g$ to get to $X'$, then follows the edge $f'$ to get to $C$).

Since $f' \circ g = f'g \neq f$, this gives the example you want.