# 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 $T:\Delta \left(c\right)\to F$ and $T\prime :\Delta \left(c\prime \right)\to F$ denote cones over $F$.

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

$\begin{array}{ccccc}\Delta \left(c\right)& & \stackrel{\alpha }{⟶}& & \Delta \left(c\prime \right)\\ & T↘& & ↙T\prime & \\ & & F& & \end{array}$\array{ \Delta(c) &{}&\stackrel{\alpha}{\longrightarrow} &{}& \Delta(c') \\ {}& \mathllap{\scriptsize{T}}\searrow &{}& \swarrow\mathrlap{\scriptsize{T'}} &{} \\ {}&{}&F&{}&{} }

commutes. Note that naturality of any such $\alpha$ implies that for all $i,j\in J$, ${\alpha }_{i}={\alpha }_{j}$, so that $\alpha =\Delta \left(\varphi \right)$ 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

$\begin{array}{ccccc}c& & \stackrel{\varphi }{⟶}& & c\prime \\ & {T}_{j}↘& & ↙T{\prime }_{j}& \\ & & F\left(j\right)& & \end{array}$\array{ c &{}&\stackrel{\phi}{\longrightarrow} &{}& c' \\ {}& \mathllap{\scriptsize{T_j}}\searrow &{}& \swarrow\mathrlap{\scriptsize{T'_j}} &{} \\ {}&{}&F(j)&{}&{} }

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 $\alpha :\Delta c⇒\Delta c\prime$ 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 $\alpha :\Delta c⇒\Delta c\prime$ 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 $\alpha :\Delta \left(c\right)\to \Delta \left(c\prime \right)$, I get

$\begin{array}{ccc}c& \stackrel{{\mathrm{Id}}_{c}}{\to }& c\\ {\alpha }_{j}↓& & ↓{\alpha }_{k}\\ c\prime & \stackrel{{\mathrm{Id}}_{c\prime }}{\to }& c\prime \end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ c & \stackrel{Id_c}{\to} & c \\ \alpha_j\downarrow && \downarrow \alpha_{k} \\ c' & \stackrel{Id_{c'}}{\to} & c' } \,.

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

$\begin{array}{ccc}c& \stackrel{{\mathrm{Id}}_{c}}{\to }& c\\ \varphi ↓& & ↓\varphi \\ c\prime & \stackrel{{\mathrm{Id}}_{c\prime }}{\to }& c\prime \\ T{\prime }_{j}↓& & ↓T{\prime }_{k}\\ F\left(j\right)& \stackrel{F\left(f\right)}{\to }& F\left(k\right)\end{array}$\array{ c & \stackrel{Id_c}{\to} & c \\ \mathllap{\scriptsize{\phi}}\downarrow && \downarrow\mathrlap{\scriptsize{\phi}} \\ c' & \stackrel{Id_{c'}}{\to} & c' \\ \mathllap{\scriptsize{T'_j}}\downarrow && \downarrow\mathrlap{\scriptsize{T'_k}} \\ F(j)&\stackrel{F(f)}{\to}&F(k) }

Revised on November 5, 2009 16:29:43 by Eric Forgy (65.163.59.49)