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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.

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    $\begingroup$ 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. $\endgroup$
    – Mark Kamsma
    Apr 11, 2019 at 9:41
  • $\begingroup$ @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. $\endgroup$
    – Daniel Donnelly
    Apr 11, 2019 at 22:57

1 Answer 1

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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.

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