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In a simplicial set, if $r<s$, does it follow that the face maps satisfy $$d_s\circ d_r = d_{r+1}\circ d_s?$$ Motivation: I want to show that for all $i<j$, $$d_{n-i}\circ d_{n-j}=d_{n-j+1}\circ d_{n-i},$$ in order to show that for each simplicial set (given by face maps $d_\bullet$, say) the opposite simplicial set (given by face maps $d_{n-\bullet}$) satisfies the simplicial identities too.

user1005113
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Imagine a triangle. Is it true that $d_1\circ d_0 = d_{1}\circ d_1$? Well, $d_1\circ d_1 ([0,1,2]) = d_1 ([0,2]) = [0],$ but $d_0([0,1,2]) = [1,2]$ already does not contain $0.$

Арсений Кряжев
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  • Thanks! But why doesn't that contradict the fact that for each simplicial set $S_\bullet$ its opposite simplicial set (see https://kerodon.net/tag/003P) $S_\bullet^\mathrm{op}$ is a simplicial set (satisfies the simplicial identities)? Note that $$(d_ i: S^{\operatorname{op}}_ n \rightarrow S^{\operatorname{op}}{n-1}) = (d{n-i}: S_ n \rightarrow S_{n-1}).$$ – user1005113 Dec 22 '21 at 12:54
  • This means that in particular we need to show $$d_{n-i}\circ d_{n-j}=d_{n-j+1}\circ d_{n-i}$$ whenever $i<j$. – user1005113 Dec 22 '21 at 12:55
  • If that's true, then this implies that $$d_s\circ d_r = d_{r+1}\circ d_s$$ if $r< s$: just plug in $i:=n-s$ and $j:=n-r$. – user1005113 Dec 22 '21 at 13:00
  • @user1005113 Note that $n$-s are different for consecutive $d$. For the second $d$ you use $n-1$ instead and then you get the usual identities. – Арсений Кряжев Dec 22 '21 at 20:04
  • What do you mean by "Note that n-s are different for consecutive d."? n-s doesn't depend on d. d isn't a variable here. – user1005113 Dec 23 '21 at 11:51
  • What do you mean by "For the second d you use n−1 instead"? What is the second d? What does it mean to "use" n-1? – user1005113 Dec 23 '21 at 11:51
  • $user1005113$ There is a composition of face maps, $d_i,$ so there is the first one and the second. In the formula $d^{op}i = d{n-i},$ $n$ does depend on $d_i$ in the sense that $n = dim(dom(d_i)).$ – Арсений Кряжев Dec 23 '21 at 21:44
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    Ah I see, thank you so much!! That helps a lot! – user1005113 Dec 27 '21 at 13:52