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In Paul Goerss and John Jardine's book on simplicial homotopy theory they define

the nerve $BC$ of a category $C$ as simplicial set determined by $BC([n])=hom_{cat}([n],C)$.

They then say in other words an $n$- simplex (by definition an element of $hom_{cat}([n],C)$) is a string of composable arrows of length $n$ in $C$.

I don't thing these are equivalent because there are more strings of composable arrows than there are elements in the hom-set:

Take for instance $C=\mathbb{Z}/5$ where $\mathbb{Z}/5$ is the category with one object and one morphism for each element in the group $\mathbb{Z}/5$. Every functor $F$ from the category $[n]$ to the category $\mathbb{Z}/5$ is determined by what $F(0\leq 1 \leq 2... \leq n)$ is. $F(0 \leq 1 \leq 2 ... \leq n )$ can be one of $5$ different arrows in the category $\mathbb{Z}/5$. Therefore there are $5$ different elements in $hom_{cat}([n], \mathbb{Z}/5)$.

The number of composable arrows of length $n$ is $5^n$.

What is wrong?

1 Answers 1

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Every functor $F$ from the category $[n]$ to the category $\mathbb{Z}/5$ is determined by what $F(0\leq 1 \leq 2... \leq n)$ is

This is not true. The morphism $0\leq n$ is just one morphism in the category $[n]$, which does not generate the entire category (if $n>1$). For instance, knowing what $F(0\leq n)$ is does not determine what $F(0\leq 1)$ is.

Knowing where $F$ sends the entire diagram $0\leq 1\leq 2\leq\dots\leq n$ does determine $F$, but that diagram does not consist of just a single map: it has a map $0\to 1$, a map $1\to 2$, and so on. These are exactly a string of $n$ composable arrows, and so give a bijection between such functors $F$ and strings of $n$ composable arrows.

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    Isn't $0 \leq 1 \leq 2 ...\leq n$ is a morphism between the objects $0$ and $n$ in $[n]$? – 2017-01-23
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    If you compose all those morphisms together, you get a single morphism from $0$ to $n$. But when I say "the entire diagram", I mean not just that single morphism but each of the individual morphisms $0\leq 1$, $1\leq 2$, and so on. – 2017-01-23
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    First, thanks for spending the time clearing my misconceptions. I do appreciate it. My understanding was that the morphisms of $[n]$ were mutually distinct paths like $0 \leq 1$, $n-1 \leq n$, $0 \leq n$, $0 \leq 1 ....\leq n$ that are not determined by what their starting and ending points are. In other words, your 'diagram' is viewed itself as a morphism in the category $[n]$. Do you have a different definition? – 2017-01-23
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    For each pair of integers $i$ and $j$ such that $0\leq i\leq j\leq n$, there is a unique morphism $i\to j$ in $[n]$. You can call this morphism whatever you want; if you want to write it as $i\leq i+1\leq\dots\leq j$ that's fine. But in any case, it is not true that the unique morphism $0\to n$ generates the entire category. – 2017-01-23
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    It seems that you are doing the following: Define the morphisms of $[n]$ to be arrows $i \to j$ such that $0 \leq i\leq j \leq n$. Functors on $[n]$ are entirely determined by what they do on $0 \to 1$, $1 \to 2$... $n-1 \to n$. One develops the theory by saying that $d^i$ is the functor such that $d^i(i-1 \to i)= i-1 \to i+1$, and similarly for $s^i$. – 2017-01-23
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    Now it is clear that $Hom([n], \mathbb{Z}/5)$ is in correspondence with a set of cardinality the number of ways of distributing $n$ pieces of gold to $5$ pirates. To be satisfied, I still need to see if the realization of this simplicial set with these $n-$ simplices for $\mathbb{Z}/2$ is the milnor space model of $B(\mathbb{Z}/2)$. – 2017-01-23