Fundamental group of simplicial space

Question

This question asks the same as mine, but it was unsuccessful in getting an answer, so I try again.

For context I am reading Weibel's K-book and am struggling with proposition 8.4 which computes $K_0$ for Waldhausen categories. Suppose we have a simplicial space $X_\bullet$ with $X_0$ point, then how can I find the fundamental group?

According to Weibel it is generated by path components of $X_1$ subject to the relations $\partial_1([y])=\partial_0([y])\partial_2([y])$ for all $[y]\in\pi_0(X_2)$. I find it completely counter intuitive that we don't actually care bout the topology of $X_1, X_2$ to compute $\pi_1$, the fact that it is very hard to visualize geometric realization of simplicial spaces is not helping me to try and get an intuition.

Any help is appreciated, thank you very much.

Answer 1

First note that $|X|$ is actually connected (so that we do not have to worry about basepoints). Namely, for a general simplicial space $X$, we have a surjection $\pi_0X_0\to\pi_0|X|$. You can either check this by direct inspection, or by a fancy homotopy colimit argument: $|X|$ is the homotopy colimit $\mathrm{hocolim}_\mathrm{[n]\in\Delta^\mathrm{op}}\,X_n$, and $\pi_0$ sends homotopy colimits of spaces to colimits of sets (it has an underlying left adjoint $\infty$-functor $\pi_0\colon\mathsf{Spaces}\to\mathsf{Set}$), so we can write $\pi_0|X|\cong\mathrm{colim}_{[n]\in\Delta^\mathrm{op}}\,\pi_0X_n$. The subcategory $1\rightrightarrows 0$ is cofinal in $\Delta^\mathrm{op}$, so $\pi_0|X|\cong\mathrm{coeq}(\pi_0X_1\rightrightarrows\pi_0X_0)$, and $\pi_0X_0$ clearly surjects onto this coequalizer. In particular, if $X_0$ is connected, so is $|X|$.

Now, let's compute $\pi_1|X|$. We use the skeleton filtration $\mathrm{sk}^0X\to\mathrm{sk}^1X\to\ldots\to\mathrm{sk}^nX\to\ldots$ of $X$, which satisfies that $X\cong\mathrm{colim}_n\,\mathrm{sk}^nX$. We can build each $\mathrm{sk}^nX$ as follows: we simply take $\mathrm{sk}^0X\cong\mathrm{const}\,X_0$ to be the constant simplicial space on $X_0$. Write, given a space $A$, also $A$ for such a constant simplicial space on $A$. Given a simplicial set $Z$, we write $Z^t$ for the simplicial space $[n]\mapsto Z_n$, where $Z_n$ is considered a discrete space. We define $\mathrm{sk}^nX$ for $n\geq 1$ via a pushout square $$\require{AMScd}\begin{CD} \partial\Delta^{n,t}\times X_n @>{\mathrm{faces}}>>\mathrm{sk}^{n-1}X\\ @VVV @VVV\\ \Delta^{n,t}\times X_n @>>>\mathrm{sk}^nX \end{CD}$$ of simplicial spaces (so the object $X_n$ is also considered a constant simplicial space). Here, the map $\mathrm{faces}$ acts as follows. Write $d^i\Delta^n$ for the $i$-th face of $\Delta^n$, which is a subspace of $\partial\Delta^n$. The map $\mathrm{faces}$, restricted to $d^i\Delta^{n,t}\times X_n$, acts as the composite $$ d^i\Delta^{n,t}\times X_n\xrightarrow{\mathrm{id}\times d_i}d^i\Delta^{n,t}\times X_{n-1}\cong \Delta^{n-1,t}\times X_{n-1}\to\mathrm{sk}^{n-1}X, $$ where the last map comes (inductively) from the pushout definition of $\mathrm{sk}^{n-1}X$.

Hitting everything in sight with the geometric realization (which preserves pushouts, and finite products of constant simplicial spaces), we hence find pushout diagrams $$\require{AMScd}\begin{CD} |\partial\Delta^{n}|\times |X_n| @>{\mathrm{faces}}>>|\mathrm{sk}^{n-1}X|\\ @VVV @VVV\\ |\Delta^{n}|\times |X_n| @>>>|\mathrm{sk}^nX| \end{CD}$$ (We could use that $|X_n|\simeq X_n$, $|\Delta^{n,t}|\simeq \Delta^n$ and $|\partial\Delta^{n,t}|\simeq\partial\Delta^n$, as the geometric realization of constant simplicial spaces $A$, or of simplicial spaces of the form $Z^t$ for $Z\in\mathsf{sSet}$, are weakly equivalent to $A$ and $Z$, respectively. Since our diagrams are homotopy pushouts anyway (all objects are cofibrant, and the left vertical map is an inclusion, and hence a cofibration of either bisimplicial sets or simplicial sets), we can safely replace objects by weakly equivalent ones and work with homotopy pushouts for the rest of the argument. I will not do so yet.)

Moreover, $\mathrm{colim}_n|\mathrm{sk}^nX|\cong|X|$ still because these colimits are also preserved. This is a sequential colimit along inclusions (cofibrations of simplicial sets), and so $\pi_1$ commutes with this colimit: we have $\pi_1|X|\cong\mathrm{colim}\,\pi_1|\mathrm{sk}^nX|$. We will compute the right hand side.

It doesn't matter what happens with $\pi_1|\mathrm{sk}^nX|$ if $n=0$ since we will take a sequential colimit anyway. For $n=1$, we have the pushout $$\require{AMScd}\begin{CD} |\partial\Delta^{1}|\times |X_1| @>{\mathrm{faces}}>>|\mathrm{sk}^{0}X|\\ @VVV @VVV\\ |\Delta^{1}|\times |X_1| @>>>|\mathrm{sk}^1X| \end{CD}$$ Now I will use $|Z^t|\simeq Z$ and $|A|\simeq A$ for $A,Z\in\mathsf{sSet}$. Together with $|\mathrm{sk}^0X|=|X_0|\simeq *$, we can rewrite the above diagram, and find that $|\mathrm{sk}^1X|$ is up to weak equivalence also computed by the homotopy pushout square $$ \require{AMScd}\begin{CD} \partial\Delta^1\times X_1 @>>> *\\ @VVV @VVV\\ \Delta^1\times X_1 @>>> |\mathrm{sk}^1X|\end{CD} $$ Since the left vertical map is an inclusion, we can compute the homotopy pushout of the ''square minus bottom right corner'' by the usual pushout of simplicial sets (i.e. the homotopy type of this strict pullback is the homotopy type of the homotopy pushout). But this strict pushout is just the reduced suspension $\Sigma (X_{1})_+$, where $(X_{1})_+$ is $X_1\sqcup *$. Therefore we find $|\mathrm{sk}^1X|\simeq\Sigma(X_{1,+})$. In particular, $\pi_1|\mathrm{sk}^1X|\cong\pi_1\Sigma(X_{1,+})\cong\pi_0\Omega\Sigma(X_{1})_+$. This is isomorphic to the free group $F(\pi_0X_1)$ on the set $\pi_0(X_1)$. There are two ways to see this: concretely, given a path in $\Sigma(X_1)_+$ from the south pole to the north pole (which are identified, hence this represents a class in $\pi_1\Sigma(X_1)_+$) that is of the form $t\mapsto[t,x]$ for $t\in[0,1]$ and $x\in X_1$ a fixed point, the class of this path in the fundamental group $\pi_1\Sigma(X_1)_+$ corresponds to the generator of $F(\pi_0X_1)$ corresponding to the path component of $x\in X_1$. You can via some more algebraic topology convince yourself that this defines the desired isomorphism. You can also be fancy and note that $\Omega\Sigma(X_1)_+$ is the free $\mathbb{E}_1$-group on $X_1$, i.e. the free homotopy coherent analogue of a group on the given space $X_1$. This free $\mathbb{E}_1$-group construction is a left adjoint $\infty$-functor, and via some abstract nonsense we conclude that $\pi_0\Omega\Sigma(-)_+$ is actually the (ordinary) free group construction on the set $\pi_0(-)$. In any case, we concluded that $\pi_1|\mathrm{sk}^1X|\cong F(\pi_0X)$ in a specific way.

Now let us look at the case $n=2$. The pushout square reads $$\require{AMScd}\begin{CD} |\partial\Delta^{2}|\times |X_2| @>{\mathrm{faces}}>>|\mathrm{sk}^{1}X|\\ @VVV @VVV\\ |\Delta^{2}|\times |X_2| @>>>|\mathrm{sk}^2X| \end{CD}$$ and again, up to weak homotopy equivalence we may write $|\mathrm{sk}^2X|$ as pushout of the diagram $$\require{AMScd}\begin{CD} \partial\Delta^{2}\times X_2 @>{\mathrm{faces}}>>\Sigma(X_1)_+\\ @VVV @VVV\\ \Delta^{2}\times X_2 @>>>|\mathrm{sk}^2X| \end{CD}$$ Given $y\in\pi_0X_2\cong\pi_0(\Delta^2\times X_2)$, write $(X_2)_y\subseteq X_2$ for the path component of $y$. We can attach the cells $\Delta^2\times (X_2)_y$ sequentially (by picking a total order on $\pi_0 X_2$). Any pushout diagram $$\require{AMScd}\begin{CD} \partial\Delta^{2}\times (X_2)_y @>{\mathrm{faces}}>>\Sigma(X_1)_+\\ @VVV @VVV\\ \Delta^{2}\times (X_2)_y @>>>P(y) \end{CD}$$ that we encounter satisifies that all objects are connected, so we can apply Seifert-Van Kampen to compute $\pi_1$. The left vertical map becomes, after applying $\pi_1$, the projection map $\mathbb{Z}\times\pi_1(X_2)_y\to\pi_1(X_2)_y$. The top horizontal map $\mathbb{Z}\times\pi_1(X_2)_y\to\pi_1\Sigma(X_1)_+\cong F(\pi_0X_1)$ will send the element $(1,e)$ (with $1\in\mathbb{Z}$ and $e\in\pi_1(X_2)_y$ the neutral element) to the element $[\partial_1 y]^{-1}*[\partial_0 y]*[\partial_2 y]$, with $\partial_i\colon X_2\to X_1$ the simplicial structure maps that $X$ has. This follows from the description of the map $\mathrm{faces}$ below our first pushout diagram, and the explicit isomorphism $\pi_1\Sigma(X_1)_+\cong F(\pi_0X_1)$ that we described above. Hence we find $\pi_1(P(y))$ to be isomorphic to $F(\pi_0X_1)$ modulo the relation that $[\partial_1 y]^{-1}*[\partial_0 y]*[\partial_2 y]=e$, where $e\in F(\pi_0X_1)$ is the neutral element.

So, for each connected component $y\in\pi_0X_2$ of $X_2$, attaching that specific connected component to $\Sigma(X_1)_+$ has on the fundamental group the effect of quotienting out from $F(\pi_0X_1)$ the relation $[\partial_1 y]=[\partial_0 y]*[\partial_2 y]$. Therefore, attaching all the connected components to $\Sigma(X_1)_+$ has the effect of quotienting out all relations $[\partial_1 y]=[\partial_0 y]*[\partial_2 y]$ for $y\in\pi_0X_2$. Therefore, we finally arrive at the conclusion that $\pi_1|\mathrm{sk}^2X|$ is isomorphic to the group you have described in your question.

We now show that the map $\pi_1|\mathrm{sk}^{n-1}X|\to\pi_1|\mathrm{sk}^{n}X|$ is an isomorphism if $n\geq 3$. This is actually also just Seifert-Van Kampen: the argument is similar to what we did above for $n=2$ (including reducing to connected components $(X_n)_y$ of $X_n$), but now we will end up in the upper left corner a term $\partial\Delta^{n}\times (X_n)_y$. Taking the fundamental group, the left hand vertical map just becomes the identity morphism $\pi_1(X_n)_y\to\pi_1(X_n)_y$ if $n\geq 3$, since $\pi_1\partial\Delta^{n-1}\cong\{e\}$ if $n\geq 3$. But in light of Seifert-Van Kampen, this in turn means that we do not actually quotient out any more relations from the fundamental group $\pi_1|\mathrm{sk}^{n-1}X|$ in order to get $\pi_1|\mathrm{sk}^{n}X|$.

Finally, we have shown that $\pi_1|X|\cong\mathrm{colim}_n\pi_1|\mathrm{sk}^{n}X|\cong\pi_1|\mathrm{sk}^{2}X|$, and that the latter group is isomorphic to the group you described in the question. With that, we are done.

Answer 2

An element of $\pi_1$ is trivial if it extends to a disk. Note that Weibel's definition basically just identifies two paths namely $\partial_0$ and $\partial 2$ if there is a $2$-cell joining them (homotopy) which in geometric realization is a an element of $X_2$. This is in $\pi_1$ if you contract a spanning tree after geometric realization.

Write $[n] \mapsto N_\bullet wS_n \mathcal{C}$ for the simplicial set in question. Note that the topology here matters because this is a simplical space by composing with the geometric realization functor.

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to get a sense for $K_0$ we want to show how to get from objects to guys in $\pi_1$.

$S_0$ is a trivial category, so we are connected.

geometric realization commutes with products so literally take $(N_\bullet wS_1 \mathcal{C}) \times I$. Note that $S_1 \mathcal{C}=\mathcal{C}$ and $I= |\Delta^1|$.

Now $S_0$ is the trivial map so collapses $0 \times \Delta^1$ and $d_0,d_1$ collapses $N_\bullet w \mathcal{C} \times \partial \Delta^1$.

the previous two guys are reduced suspension and through inclusion we get a map

$\Sigma|N_\bullet w \mathcal{C}| \to |N_\bullet w S_\bullet \mathcal{C}|$. adjointing over is the map to $\Omega |w S_\bullet \mathcal{C}|$. One needs to check the $\pi_0$ condition now.

Every object in $\mathcal{C}$ determines an object of $\pi_1|wS_\bullet \mathcal{C}|$. Now you gotta check relations in $\pi_0$ of the nerve of $S_2$

(see 8.4.6 here and the pages before it for a very concrete (although with some holes to fill) description for how this story goes.

Answer 3

Do you know that the fundamental group of a CW complex depends only on its 2-skeleton? Then it will no longer be surprising.

There are a few ways to see this. I really like the covering-theoretic exposition here which uses a simplicial analogue of the classical theory to demonstrate what you say. Also consult the Kerodon and the text of Goerss and Jardine, "Simplicial Homotopy Theory" for the relevant and supporting material.

I realise this is a brief answer but let me know if you're struggling to find a specific proof within these texts or are hung up on one of the details. I mean, all of these texts omit (many) details, as is the norm, so there is always room for confusion and for the reader to do their own work.