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For pointed simplicial sets there are two equivalent definitions of the basepoint. Let $\Delta^0$ be the simplicial set with only one vertex in each degree. Let $X$ be a simplicial set. Then a basepoint in $X$ is either a simplicial map $\varphi:\Delta^0\to X$ or a distinguished point $\ast\in X_0$. I see that the map $\varphi$ specifies a point in $X_0$ (and any $X_i$ for $i>0$).

My question: How does, on the other hand, a point $\ast\in K_0$ determine a map $\varphi:\Delta^0\to X$? How are these two notions equivalent?

Mike
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1 Answers1

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$\newcommand{\De}{\Delta}$ $\newcommand{\Set}{\text{Set}}$ This is all very much representable functors and Yoneda lemma stuff.

A simplicial set $X$ is a contravariant functor from the simplex category $\De$ to $\Set$. So for each object $[n]$ of $\De$ we have a set $X_n$ and also appropriate maps between these. Simplicial sets form a category $\hat\De=\Set^{\De^{\text{op}}}$. Yoneda's lemma shows that there is an embedding $\De\to\hat\De$ where $[n]$ is mapped to the simplicial set $\De^n$ defined by $\De^n:[m]\mapsto \text{Mor}_\De([m],[n])$. For each $n$ there is a functor from $\hat\De$ to $\Set$ given by $X\mapsto X_n$. Yoneda's lemma tells us this functor is represented by $\De^n$, that is $X_n\cong \text{Map}_{\hat\De}(\De^n,X)$.

Your question about pointing is the case $n=0$ looked at in these two ways. Picking a point in $X_0$ is picking a point in $X_0$, but that is naturally equivalent to picking a simplicial map from $\De^0$ to $X$; this is the equivalence $X_0\cong \text{Map}_{\hat\De}(\De^0,X)$.

Angina Seng
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