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In the answer given by @Kevin Arlin in the MSE question https://math.stackexchange.com/a/2994934/820022 if I am not mistaken the geometric realization of a simplicial set $X$ is defined as a colimit of the following diagram in Top :

$\pi \circ p:\Delta \downarrow X \rightarrow Top$ where $\pi:\Delta \rightarrow Top$ is defined as the canonical functor sending $[n] \rightarrow |\Delta^n|$ in object level (where $|\Delta^n|$ is the standard geometric $n$-simpplex) and also appropriately defined in the morphism level whereas I guess $p:\Delta \downarrow X \rightarrow \Delta$ is defined as follows:

On Objects: $(\sigma:\Delta^n \rightarrow X) \mapsto [n] \in \Delta,$ the usual finite ordinal category.

On Morphisms: enter image description here $\theta \mapsto (\theta_{*}:[n] \rightarrow [m])$

where $\theta_{*}:[n] \rightarrow [m]$ is defined as $\theta_{[n]}(1_{[n]})$ (Coming from contravariant Yoneda lemma).

I was trying to show that $p$ is indeed a functor but I am struck at the following step while showing $(\psi \circ \phi)_{*}= \psi_{*} \circ \phi_{*}$ where $\psi:\Delta^m \rightarrow \Delta^{r}$ and $\phi: \Delta^{n} \rightarrow \Delta^{m}$. (Though I made abuse of notation for convenience).

My confusion:

$\psi_{*} \circ \phi_{*}= \psi_{[m]}(1_{[m]}) \circ \phi_{[n]}(1_{[n]})$......(1)

$(\psi \circ \phi)_{*}=(\psi \circ \phi)_{[n]}(1_{[n]})= \psi_{[n]} \circ \phi_{[n]}(1_{[n]})$.....(2)

I am not able to show (1) = (2).

Am I misunderstanding anything?

Or did I interpret the answer by @Kevin Arlin in an incorrect way?

Thanks in advance.

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    $\begingroup$ What does the Yoneda lemma tell you about where $\psi$ sends $\phi(1_{[n]})$? Also your question might also be phrased as "Why is the Yoneda embedding functorial"? $\endgroup$
    – jgon
    Aug 31, 2020 at 15:41
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    $\begingroup$ Sure, but the Yoneda embedding is an embedding, so it being functorial means the inverse is functorial as well. $\endgroup$
    – jgon
    Aug 31, 2020 at 16:12
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    $\begingroup$ @jgon Thanks I got it $\endgroup$ Aug 31, 2020 at 16:19
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    $\begingroup$ Sorry, I've been being brief, since I'm on mobile. What I mean is that if $F$ is a faithful functor, and $F(f)F(g)=F(h)$, then $F(fg)=F(h)$, so $h=fg$ by faithfulness. We don't need the full strength of an embedding for this result, or an inverse equivalence. In our case we know that the original maps are the Yoneda embedding applied to the starred versions. $\endgroup$
    – jgon
    Aug 31, 2020 at 16:20
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    $\begingroup$ Glad to see you worked it out. $\endgroup$
    – jgon
    Aug 31, 2020 at 16:21

1 Answer 1

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Let $Y:\Delta \rightarrow sSets$ be the Yoneda embedding. By Yoneda embedding there exist $g:[m] \rightarrow [r]$ and $f:[n] \rightarrow [m]$ such that $\psi = Y(g)$ and $\phi= Y(f)$.

So $\psi \circ \phi=Y(g) \circ Y(f)$.

Now component wise, $(\psi \circ \phi)_{[k]}= \psi_{[k]} \circ \phi_{[k]}= Y(g)_{[k]} \circ Y(f)_{[k]}$ for $[k] \in \Delta$.

So, in particular $\psi_{[m]}(1_{[m]}) \circ \phi_{[n]}(1_{[n]})=Y(g)_{[m]}(1_{[m]}) \circ Y(f)_{[n]}(1_{[n]})= g \circ 1_{[m]} \circ f \circ 1_{[n]}=g \circ f= \psi_{*} \circ \phi_{*}$ (using the same notation as mentioned in the question.)

On the other hand, $(\psi \circ \phi)_{*}= (\psi \circ \phi)_{[n]}(1_{[n]})= (Y(g) \circ Y(f))_{[n]}(1_{[n]})=Y(g \circ f)_{[n]}(1_{[n]})= g \circ f \circ 1_{[n]}= g \circ f$.

Hence $\psi_{*} \circ \phi_{*}= (\psi \circ \phi)_{*}$.

So, $p:\Delta \downarrow X \rightarrow \Delta$ is a functor. (Proved)

(Indentity Preservation is easy to show.)

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    $\begingroup$ Yeah that works, although the Yoneda lemma also says that $\psi=Y(\psi_*)$, so you don't really need to introduce new morphisms $f$ and $g$. However, this argument is correct. $\endgroup$
    – jgon
    Aug 31, 2020 at 19:03
  • $\begingroup$ @jgon Thank you I got your point. $\endgroup$ Aug 31, 2020 at 19:28

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