nLab
cone morphism

Definition

Let $F:J\to C$ be a diagram in a category $C$. Also, for any objects $c,c\prime$ in $C$, let $T:\Delta (c)\to F$ and $T\prime :\Delta (c\prime )\to F$ denote cones over $F$.

A cone morphism is a natural transformation $\alpha :\Delta (c)\to \Delta (c\prime )$ such that the diagram

commutes. Note that naturality of any such $\alpha$ implies that for all $i,j\in J$, ${\alpha }_i={\alpha }_j$, so that $\alpha =\Delta (\varphi )$ for some $\varphi :c\to c\prime$ in $C$. The single component $\varphi$ itself is often referred to as the cone morphism.

An equivalent definition of a cone morphism $\varphi :T\to T\prime$ says that all component diagrams

commute.

Discussion

The following discussion took place at the component diagram above:

Eric: What is the “component free” way to say that?

Finn Lawler: I think the category of cones over $F$ is the comma category $\Delta /F$, so that a morphism $\alpha :T\to T\prime$ should be just a natural transformation $\alpha :\Delta c\Rightarrow \Delta c\prime$ such that $T\prime \alpha =T$. That gives your condition in components, I think.

Eric: Thanks Finn! I’m still learning all this, so it’ll take me some time to absorb what you said. It sounds good though :) Either way, it sounds like some potentially good additional content.

Finn Lawler: I should point out that a natural transformation $\alpha :\Delta c\Rightarrow \Delta c\prime$ is very nearly exactly the same thing as a morphism $\varphi :c\to c\prime$ (it’s $\varphi$ in each component, which you’ll see if you draw $\alpha$’s naturality square, so it’s $\Delta \varphi$ for some $\varphi$). Now look at the triangle above, write $\Delta \varphi :\Delta c\to \Delta c\prime$ instead of $\varphi$ and erase the $j$s and you have the morphism in the comma category.

I hope this helps. If it’s done the opposite, apologies. I’ve a habit of trying the one and accomplishing the other.

Eric: Hmm. I’m probably confused, but when I draw the naturality square for $\alpha :\Delta (c)\to \Delta (c\prime )$, I get

for every $j,k\in J$.

Eric: I think I got it. My diagram is correct, except we have ${\alpha }_j={\alpha }_k$ and we want this to be $\varphi :c\to c\prime$. I made that more explicit in the definition above by adding “whose component is $\varphi :c\to c\prime$.”

The full blown diagram looks like