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

Question

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 $\tilde{\Delta}$?

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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.

Answer 1

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.