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1. • CommentRowNumber1.
   • CommentAuthorporton
   • CommentTimeSep 20th 2015
   • PermaLink
   Author: porton
   Format: MarkdownItex[ Higher Operads, Higher Categories](http://arxiv.org/abs/math/0305049), page 47 says that a $\mathbf{Set}$-graph is an ordinary directed graph. I don't understand this: $\mathbf{Set}$-graph is an $X_0\times X_0$-indexed family of objects of $\mathbf{Set}$, that is a family of sets. Thus edges of the graph are sets. It makes no sense for me that edges are sets. Moreover, edges of "an ordinary directed graph" are not necessarily sets (unless we assume something like standard ZF where every object is a set).

   Higher Operads, Higher Categories, page 47 says that a $\mathbf{Set}$-graph is an ordinary directed graph.

   I don’t understand this: $\mathbf{Set}$-graph is an $X_0\times X_0$-indexed family of objects of $\mathbf{Set}$, that is a family of sets. Thus edges of the graph are sets.

   It makes no sense for me that edges are sets. Moreover, edges of “an ordinary directed graph” are not necessarily sets (unless we assume something like standard ZF where every object is a set).

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 20th 2015
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItex> Thus edges of the graph are sets. No, that's not the right idea. The underlying idea is something that's well worth getting used to: that a function $f: Y \to X$ can be regarded equivalently as a family of sets $Y_x = f^{-1}(\{x\})$. (Note that the $Y_x$ could be empty.) Thus, given a directed graph consisting of a set $X_0$, a set $E$, and a source-target function $\langle d_0, d_1 \rangle: E \to X_0 \times X_0$, we can assign to each pair of vertices $(x, y) \in X_0 \times X_0$ the *set* of edges $e$ such that $d_0(e) = x$ and $d_0(e) = y$. Conversely, given a family $\{F_{(x, y)}\}_{(x, y) \in X_0 \times X_0}$, form a directed graph where the edge set $E$ is the disjoint union $\bigsqcup_{(x, y) \in X_0 \times X_0} F_{(x, y)}$, and the source-target pairing satisfies $\langle d_0, d_1 \rangle(e) = (x, y)$ for $e \in E$ precisely when $e \in F_{(x, y)}$. So directed graphs are mathematically equivalent to $\mathbf{Set}$-graphs.

   Thus edges of the graph are sets.

   No, that’s not the right idea.

   The underlying idea is something that’s well worth getting used to: that a function $f: Y \to X$ can be regarded equivalently as a family of sets $Y_x = f^{-1}(\{x\})$. (Note that the $Y_x$ could be empty.)

   Thus, given a directed graph consisting of a set $X_0$, a set $E$, and a source-target function $\langle d_0, d_1 \rangle: E \to X_0 \times X_0$, we can assign to each pair of vertices $(x, y) \in X_0 \times X_0$ the set of edges $e$ such that $d_0(e) = x$ and $d_0(e) = y$. Conversely, given a family $\{F_{(x, y)}\}_{(x, y) \in X_0 \times X_0}$, form a directed graph where the edge set $E$ is the disjoint union $\bigsqcup_{(x, y) \in X_0 \times X_0} F_{(x, y)}$, and the source-target pairing satisfies $\langle d_0, d_1 \rangle(e) = (x, y)$ for $e \in E$ precisely when $e \in F_{(x, y)}$.

   So directed graphs are mathematically equivalent to $\mathbf{Set}$-graphs.

3. • CommentRowNumber3.
   • CommentAuthorporton
   • CommentTimeSep 20th 2015
   • (edited Sep 20th 2015)
   • PermaLink
   Author: porton
   Format: MarkdownItex@Todd_Trimble What you described above is completely clear. I have no trouble understanding this. Don't count me stupid. The problem that $F_{(x,y)}$ in "Higher Operads, Higher Categories" is a set of objects, not a set of morphisms. The trouble is that in that book $X(x,x^')$ (which is considered the set of edges of the constructed graph) are **objects** of $\mathcal{V}$ not morphisms. For me it makes no sense to consider objects as edges. So I don't understand how $\mathbf{Set}$-graphs are equivalent to directed graphs. Sorry. It seems easy but I don't understand. Maybe, it is a mere typo and we should read "of morphisms of $\mathcal{V}$" instead "of objects of $\mathcal{V}$" at page 47?

   @Todd_Trimble What you described above is completely clear. I have no trouble understanding this. Don’t count me stupid.

   The problem that $F_{(x,y)}$ in “Higher Operads, Higher Categories” is a set of objects, not a set of morphisms.

   The trouble is that in that book $X(x,x^')$ (which is considered the set of edges of the constructed graph) are objects of $\mathcal{V}$ not morphisms. For me it makes no sense to consider objects as edges.

   So I don’t understand how $\mathbf{Set}$-graphs are equivalent to directed graphs. Sorry. It seems easy but I don’t understand.

   Maybe, it is a mere typo and we should read “of morphisms of $\mathcal{V}$” instead “of objects of $\mathcal{V}$” at page 47?

4. • CommentRowNumber4.
   • CommentAuthorporton
   • CommentTimeSep 20th 2015
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   Author: porton
   Format: MarkdownItexOh, I've understood my error: "A family" here means an unordered set. I confused it with an indexed family. Now it seems clear.

   Oh, I’ve understood my error:

   “A family” here means an unordered set. I confused it with an indexed family.

   Now it seems clear.

5. • CommentRowNumber5.
   • CommentAuthorporton
   • CommentTimeSep 20th 2015
   • PermaLink
   Author: porton
   Format: MarkdownItexWrong, nevermind my previous comment. Now I have really understood. Issue closed. Sorry for your time taken by my stupidity.

   Wrong, nevermind my previous comment.

   Now I have really understood. Issue closed. Sorry for your time taken by my stupidity.