# How to see that $S(\sigma) = \text{Hom}_{\Delta}(\sigma, [0,1])$ maps to the category $\tilde{\Delta}^{\text{op}}$.

Let $$\Delta$$ be the simplicial category. Let $$\tilde{\Delta}$$ be the subcategory of non-empty totally ordered sets as objects and order-preserving maps that also preserve the smallest and largest elements in totally ordered sets.

Every time I see $$\text{op}$$ used I struggle. How can you easily identify when a functor is covariant but maps to $$\tilde{\Delta}^{\text{op}}$$? Specifically there's a functor from $$\Delta \to \tilde{\Delta}^{\text{op}}$$ given by $$\sigma \mapsto \text{Hom}_{\Delta}(\sigma, [0,1])$$ where that homset has the structure of a totally ordered set given by $$f\leq g \iff f(i) \leq g(i)$$ for every $$i \in \sigma$$.

How can you see that the functor $$S$$ maps to the opposite category of $$\tilde{\Delta}$$?

If we have $$\sigma \xrightarrow{f} \tau \xrightarrow{g} \phi$$ in $$\Delta$$, then $$\text{Hom}_{\Delta}(\tau, [0, 1]) \xrightarrow{S(g)} \text{Hom}_{\Delta}(\phi, [0,1])$$ in $$\Delta^{\text{op}}$$ since $$S(g)(h) := h \circ g \in \text{Hom}_{\Delta}(\phi, [0,1])$$ for each $$h \in \text{Hom}_{\Delta}(\tau, [0,1])$$. In other words ?

I think I got that part wrong.

• There is something wrong in that last example. You are right that $S(g)(h) = h \circ g$. However, this lives in $Hom_\Delta(\tau, [0,1])$ not in $Hom_\Delta(\phi, [0,1])$. You have an arrow $g: \tau \to \phi$, so the domain of $h \circ g$ is going to be $\tau$. It might help to keep drawing arrows and their composition as you did before: $\tau \xrightarrow{g} \phi \xrightarrow{h} [0,1]$ shows you directly where everything is going and coming from. – Mark Kamsma Apr 11 at 9:41
• @MarkKamsma I have adult adhd I think. I start drawing a diagram and never finish! I love these diagrams though. Hence I made a tool to help us lower math mortals. See my profile for link. – BananaCats Category Theory App Apr 11 at 22:57

## 1 Answer

I think I have a method for those of us who suck at this.

• Suppose $$S: \Delta \to \Delta^{\text{op}}$$.
• This means that $$S : \text{Hom}_{\Delta}(\sigma, \tau) \to \text{Hom}_{\Delta^{\text{op}}}(\sigma, \tau) = \text{Hom}_{\Delta}(\tau, \sigma)$$.
• $$S(f\circ g) = S(f) \circ_{\text{op}} S(g) = S(g) \circ S(f)$$.
• Clearly we need pre-composition $$\circ f$$ in the problem.
• Say $$f : \sigma \to \tau$$ in $$\Delta$$. Then $$S(f) : S(\tau) \to S(\sigma)$$.

Now all together it should make sense.