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1. • CommentRowNumber1.
   • CommentAuthorMirco Richter
   • CommentTimeMay 14th 2012
   • (edited May 14th 2012)
   • PermaLink
   Author: Mirco Richter
   Format: MarkdownItexWould like to change the definition of the ordinal sum functor in the simplex category: $ (f\oplus g)(i) = \left\lbrace \array{ f(i) & if 0 \leq i \leq m - 1 \\ m' + g(i-m) & if m \lt i \leq (m+n-1) } \right. \,. $ must be $ (f\oplus g)(i) = \left\lbrace \array{ f(i) & if 0 \leq i \leq m - 1 \\ m' + g(i-m) & if m \leq i \leq (m+n-1) } \right. \,. $ since otherwise, the value of $(f\oplus g)(m)$ is not defined. But moreover I dont understand the following situation in the definition: Suppose $f: \mathbf{0} \to \mathbf{m}'$ is the unique morphism from the empty set into $\mathbf{m}'$ and $g:\mathbf{0} \to \mathbf{n}'$ is the unique morphism from the empty set into $\mathbf{n}'$. What is $f\oplus g$? Must be the unique morphism into $\mathbf{m}'\oplus \mathbf{n}'$ but this isn't clear from the definition of $f \oplus g$.

   Would like to change the definition of the ordinal sum functor in the simplex category:

   $(f\oplus g)(i) = \left\lbrace \array{ f(i) & if 0 \leq i \leq m - 1 \\ m' + g(i-m) & if m \lt i \leq (m+n-1) } \right. \,.$

   must be

   $(f\oplus g)(i) = \left\lbrace \array{ f(i) & if 0 \leq i \leq m - 1 \\ m' + g(i-m) & if m \leq i \leq (m+n-1) } \right. \,.$

   since otherwise, the value of $(f\oplus g)(m)$ is not defined. But moreover I dont understand the following situation in the definition:

   Suppose $f: \mathbf{0} \to \mathbf{m}'$ is the unique morphism from the empty set into $\mathbf{m}'$ and $g:\mathbf{0} \to \mathbf{n}'$ is the unique morphism from the empty set into $\mathbf{n}'$. What is $f\oplus g$? Must be the unique morphism into $\mathbf{m}'\oplus \mathbf{n}'$ but this isn’t clear from the definition of $f \oplus g$.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 14th 2012
   • (edited May 14th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexRight, I turned that "$\lt$" into a "$\leq$" in the entry. Concerning your question: if $\mathbf{m} = 0$ and $\mathbf{n} = 0$ then the range for $i$ is in both cases $0 \leq i \leq -1$. There is no such value and hence there is nothing to do and so the morphism is uniquely specified. This is the usual way of reading such case distinctions. But if you find it non-obvious, feel free to add a remark to the entry explaining this.

   Right, I turned that “$\lt$” into a “$\leq$” in the entry.

   Concerning your question: if $\mathbf{m} = 0$ and $\mathbf{n} = 0$ then the range for $i$ is in both cases $0 \leq i \leq -1$. There is no such value and hence there is nothing to do and so the morphism is uniquely specified. This is the usual way of reading such case distinctions. But if you find it non-obvious, feel free to add a remark to the entry explaining this.