cone morphism


Let F:JCF:J\to C be a diagram in a category CC. Also, for any objects c,cc,c' in CC, let T:Δ(c)FT:\Delta(c)\to F and T:Δ(c)FT':\Delta(c')\to F denote cones over FF.

A cone morphism is a natural transformation α:Δ(c)Δ(c)\alpha\colon \Delta(c)\to\Delta(c') such that the diagram

Δ(c) α Δ(c) T T F \array{ \Delta(c) &{}&\stackrel{\alpha}{\longrightarrow} &{}& \Delta(c') \\ {}& \mathllap{\scriptsize{T}}\searrow &{}& \swarrow\mathrlap{\scriptsize{T'}} &{} \\ {}&{}&F&{}&{} }

commutes. Note that naturality of any such α\alpha implies that for all i,jJi,j\in J, α i=α j\alpha_i=\alpha_j, so that α=Δ(ϕ)\alpha=\Delta(\phi) for some ϕ:cc\phi : c \to c' in CC. The single component ϕ\phi itself is often referred to as the cone morphism.

An equivalent definition of a cone morphism ϕ:TT\phi : T \to T' says that all component diagrams

c ϕ c T j T j F(j) \array{ c &{}&\stackrel{\phi}{\longrightarrow} &{}& c' \\ {}& \mathllap{\scriptsize{T_j}}\searrow &{}& \swarrow\mathrlap{\scriptsize{T'_j}} &{} \\ {}&{}&F(j)&{}&{} }



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 FF is the comma category Δ/F\Delta / F, so that a morphism α:TT\alpha : T \to T' should be just a natural transformation α:ΔcΔc\alpha : \Delta c \Rightarrow \Delta c' such that Tα=TT' \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 α:ΔcΔc\alpha: \Delta c \Rightarrow \Delta c' is very nearly exactly the same thing as a morphism ϕ:cc\phi: c \to c' (it’s ϕ\phi in each component, which you’ll see if you draw α\alpha’s naturality square, so it’s Δϕ\Delta \phi for some ϕ\phi). Now look at the triangle above, write Δϕ:ΔcΔc\Delta \phi : \Delta c \to \Delta c' instead of ϕ\phi and erase the jjs 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 α:Δ(c)Δ(c)\alpha:\Delta(c)\to\Delta(c'), I get

c Id c c α j α k c Id c c. \array{ c & \stackrel{Id_c}{\to} & c \\ \alpha_j\downarrow && \downarrow \alpha_{k} \\ c' & \stackrel{Id_{c'}}{\to} & c' } \,.

for every j,kJj,k\in J.

Eric: I think I got it. My diagram is correct, except we have α j=α k\alpha_j = \alpha_k and we want this to be ϕ:cc\phi:c\to c'. I made that more explicit in the definition above by adding “whose component is ϕ:cc\phi:c\to c'.”

The full blown diagram looks like

c Id c c ϕ ϕ c Id c c T j T k F(j) F(f) F(k) \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 (