Confusions about cone definition in category. I am new to category theory.
The definition of "cone" in Wiki:
Let $F : J \to C$ be a diagram in $C$.
Let $N$ be an object of $C$. A cone from $N$ to $F$ is a family of morphisms
$$
\psi _{X}\colon N\to F(X)\,
$$
for each object $X$ of $J$, such that for every morphism $f : X \to Y$ in $J$ the following diagram commutes:

The most confusion part is "a family of morphisms"
My questions are:

*

*If one object $Y$ of $J$, but there is no morphisms from $N$ to $F(Y)$, so
$N \to F(Y)$ simply do not exist in the cone?  Could this conclusion be changed if $f : X \to Y$ do exist in $J$ ?


*If one object $Z$ of $J$, but there are more than 1 morphisms from $N$ to $F(Z)$. So in this case, should we pick 1 morphism from all the morphisms from $N$ to $F(Z)$, according to  the commutative diagram ? But what if none of those morphisms could satisfy the commutative diagram , so none of those morphisms in the cone ?  Could this conclusion be changed if $f : X \to Z$ do exist in $J$ ?


*It seems that a cone is closely related to $J$.  Is it right to say "different $J_1, J_2$ may have different cone from $N$ ? or even did not exist a cone " ?
Thank you ! Any examples of cone is much appreciated!
 A: Let's write such a cone as $\psi : N \to F$. Notice that it's the same as a natural transformation / morphism of functors $\psi : \Delta(N) \to F$, where $\Delta(N)$ denotes the constant functor with value $N$.

*

*If there is some object $Y \in J$ such that no morphism $N \to F(Y)$ exists, this implies that there is no cone $N \to F$. Your question "Could this conclusion be changed ..." is unclear.


*Certainly there can be any number of morphisms $N \to F(Z)$. A cone just picks one of these morphisms, for every object $Z$ in the index category $J$. And not just that: the choices need to be "compatible", meaning that the mentioned diagrams commute. And yes, this is not always possible. Again, your question "Could this conclusion be changed ..." is unclear.


*Of course the whole definition depends on $J$ in a very important way. Just look at some examples and compare the notions of cones: $J = ( \bullet ~~~ \bullet)$ yields a binary product cone, $J = (\bullet \rightarrow \bullet \leftarrow \bullet)$ yields a pullback cone, $J = (~)$ yields an empty cone. I don't know if this is what you are after, but every functor $J_2 \to J_1$ induces for every cone $J_1$-shaped cone a $J_2$-shaped cone, simply by precomposing the diagram with that functor.
