[Jonathan wrote](https://forum.azimuthproject.org/profile/2316/Jonathan%20Castello):
> Given arrows \$$F(f) : F(X) \to F(Y)\$$ and \$$F(g) : F(Y) \to F(Z)\$$, we want to show that \$$F(g) \circ F(f)\$$ is well-defined and is in the range of \$$F\$$. Note that there might be multiple \$$f\$$ and \$$g\$$ giving the same arrows; for now, pick any representatives. If \$$F\$$ is injective on objects, then we know that the codomain of \$$f\$$ and the domain of \$$g\$$ are the same; so the two are compatible. So \$$g \circ f\$$ exists, and \$$F(g) \circ F(f) = F(g \circ f)\$$ is in the range of \$$F\$$.
>
> Now consider \$$f', g'\$$ distinct from \$$f, g\$$ such that \$$F(f) = F(f')\$$ and \$$F(g) = F(g')\$$. Then \$$g \circ f\$$ and \$$g' \circ f'\$$ exist, but need not be the same. Nonetheless, \$$F(g' \circ f') = F(g') \circ F(f') = F(g) \circ F(f) = F(g \circ f)\$$.
>
> This would seem to imply that no matter which representatives \$$f, g\$$ we pick, their composition gives the same image under \$$F\$$. So the range of \$$F\$$ is closed under a well-defined composition.
>
> What am I missing?

In a category, if you have a set of morphisms \$$\mathbf{Mor}(\mathcal{C})\$$, then you need if \$$f: A \to B,g: B \to C \in \mathbf{Mor}(\mathcal{C})\$$ then there must be some \$$h : A \to C \in \mathbf{Mor}(\mathcal{C})\$$ such that \$$g \circ f = h\$$.

Unfortunately, the image of a functor might not be closed under composition like that.

I have tried to make a picture of the example in that link below.

![](https://svgshare.com/i/72W.svg)

Above on the left is the original category \$$\mathcal{C}\$$, and on the right is the category \$$\mathcal{D}\$$ that the functor \$$F : \mathcal{C} \to \mathcal{D}\$$ maps to.

As you can see, if we look at the morphisms in the image of \$$F\$$ we have \$$\mathbf{Mor}(F(\mathcal{C})) = \lbrace F(f): X \to Y, F(g): Y \to Z, F(id_A): X \to X, F(id_B): Y \to Y, F(id_C): Y \to Y, F(id_D): Z \to Z \rbrace\$$.

I have also drawn the missing morphism \$$? : X \to Z\$$, where \$$? = F(g) \circ F(f)\$$.

If \$$F(\mathcal{C})\$$ was a category then we would have \$$? \in \mathbf{Mor}(F(\mathcal{C}))\$$, but it is not.

Does this help clarify what is going on?