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I'm reading Tammo tom Dieck's algebraic topology book about Seifert-van Kampen theorem (Theorem 2.6.2).

(2.6.2) Theorem (Seifert-van Kampen). Let $X_0$ and $X_1$ be subspaces of $X$ such that the interiors cover $X$. Let $i_v:X_{01}=X_0\cap X_1\to X_v$ and $j_v:X_v\to X$ be the inclusions. Suppose that $X_0,X_1,X_{01}$ are path connected with base point $\ast\in X_{01}$. Then \begin{matrix}\pi_1(X_{01},\ast) & \to & \pi_1(X_0,\ast)\\ \downarrow&&\downarrow \\ \pi_1(X_1,\ast) & \to & \pi_1(X,\ast) \end{matrix} is a pushout in the category of groups.

But the statement of Seifert-van Kampen theorem in nlab does not require $X_0$ and $X_1$ to be path connected, which says:

Theorem. Let $X$ be a topological space covered by open subsets $X_0,X_1\subset X$ such that $X_{01}=X_0\cap X_1$ is path connected. Then for any choice of base point $\ast\in X_{01}$, \begin{matrix}\pi_1(X_{01},\ast) & \to & \pi_1(X_0,\ast)\\ \downarrow&&\downarrow \\ \pi_1(X_1,\ast) & \to & \pi_1(X,\ast) \end{matrix} is a pushout in the category of groups. https://ncatlab.org/nlab/show/van+Kampen+theorem

My question: Is the the version of Seifert-van Kampen theorem in nlab correct ? If it is correct, is the the version of Seifert-van Kampen theorem in nlab a corollary of the version of Seifert-van Kampen theorem in Tammo tom Dieck's book?

I couldn't find the proof for the version of Seifert-van Kampen theorem in nlab after searching the Internet. Can anybody help me? Thanks!


EDIT: This problem still needs to be solved.

Our attempts are as following: For a pointed space $Z$ write $\tilde{Z}$ for the path-component containing the basepoint. Under the hypothesis of nlab's theorem, following the comments by Tyrone, we have $\tilde{X}=\tilde{X_0}\cup \tilde{X_1}$ and $\tilde{X_{01}}=\tilde{X_0}\cap \tilde{X_1}$. But to apply Dieck's theorem, as commented by Paul Frost, we must know that $\tilde{X}$ is covered by the interiors of $\tilde{X_0}$ and $\tilde{X_1}$ rel $\tilde{X}$. But this can not be seen easily.

The version of Seifert-van Kampen theorem in nlab still needs a proof. (Even if not as an corollary of the version in tom Dieck.) Can anybody give some reference or idea ?

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    $\begingroup$ Note that $\pi_1$ only sees the path-component of the baseboint. Thus while path-connectedness of $X_0,X_1$ is not necesseary (as per nlab's statement), you don't get anything new by dropping this assumption. $\endgroup$
    – Tyrone
    Apr 20, 2021 at 14:36
  • $\begingroup$ @Tyrone Thank you for your comment! But the path connectedness of $X_0$ and $X_1$ seems to be crucial in the proof of Seifert-van Kampen theorem in Tammo tom Dieck's book. And I am not able to deduce the version of Seifert-van Kampen theorem in nlab from the version in Tammo tom Dieck's book. Could you explain it in more detail? $\endgroup$
    – Z.H. He
    Apr 20, 2021 at 14:45
  • $\begingroup$ For a pointed space $Z$ write $\widetilde Z$ for the path-component containing the basepoint. Then $\pi_nZ=\pi_n\widetilde Z$ (more formally the inclusion $\widetilde Z\subseteq Z$ induces an isomorphism). Now replace $X_0,X_1$ by $\widetilde X_0,\widetilde X_1$ and prove SvK (assuming $X_{01}=\widetilde X_0\cap\widetilde X_1$ is path-connected). Because $\pi_1X_1=\pi_1\widetilde X_1$, etc... you now have a statement for the general case. Jeff Strom's book includes a statement of SvK without the assumption of path-connectedness of $X_0,X_1$. $\endgroup$
    – Tyrone
    Apr 20, 2021 at 15:03
  • $\begingroup$ @Tyrone Thanks for your supplement! The answer by Jackson explains your comment in more detail. But I am still confused with the equality $\tilde{X} = \tilde{X}_0 \cup \tilde{X}_1$. I can only see $\tilde{X} \supset \tilde{X}_0 \cup \tilde{X}_1$. Why $\tilde{X} \subset \tilde{X}_0 \cup \tilde{X}_1$ holds? do I miss something? $\endgroup$
    – Z.H. He
    Apr 21, 2021 at 0:03
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    $\begingroup$ To apply to Dieck's theorem it is not sufficient to know that $\tilde X = \tilde X_0 \cup \tilde X_1$. We must know that $\tilde X$ is covered by the interiors of $\tilde X_0, \tilde X_1$ rel $\tilde X$. I do not see that here (unless $X$ is locally path connected). $\endgroup$
    – Paul Frost
    Apr 22, 2021 at 13:53

2 Answers 2

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Reducing the nlab-theorem to tom Dieck's theorem breaks down when one tries to show that the interiors of $\tilde X_0, \tilde X_1$ cover $\tilde X$. At least there is no simple proof - but nevertheless it could be true. Anyway, we do not need it. In fact, tom Dieck's theorem relies on two ingredients:

  1. Theorem (2.6.1) which states a pushout property for fundamental groupoids under the assumption that $X_0$ and $X_1$ are subspaces of $X$ such that the interiors cover $X$.

  2. The existence of a retraction functor $r : \Pi(Z) \to \Pi(Z,z)$ which tom Dieck only defines for path connected $Z$. This works as follows: For each object $x$ of $\Pi(Z)$ (i.e. each point $x \in Z$) we define $r(x) = z$. For the morphisms we proceed as follows: We choose any morphism $u_x : x \to z$ if $x \ne z$ and take $u_z = id_z$= path homotopy class of the constant path at $z$. Given a morphism $\alpha : x \to y$ in $\Pi(Z)$, we define $r(\alpha) = u_y \alpha u_x^{-1}$.

We shall see that 2. can be generalized so that we can prove

Theorem (Seifert - van Kampen). Let $X$ be a topological space and $X_0,X_1\subset X$ be subsets whose interiors cover $X$ such that $X_{01}=X_0\cap X_1$ is path connected. Then for any choice of base point $\ast\in X_{01}$ \begin{matrix}\pi_1(X_{01},\ast) & \to & \pi_1(X_0,\ast)\\ \downarrow&&\downarrow \\ \pi_1(X_1,\ast) & \to & \pi_1(X,\ast) \end{matrix} is a pushout in the category of groups.

Proof. As Tyrone suggested in his comment, for a pointed space $(Z,z)$ let us denote by $\tilde Z$ the path-component of $Z$ containing the basepoint $z$. From Jackson's and your answers we know that for $X_{01}$ path connected and $* \in X_{01}$ we have $\tilde X_{01} := \tilde X_0 \cap \tilde X_1 = X_{01}$ and $\tilde X = \tilde X_0 \cup \tilde X_1$. Note that $X_{01}$ path connected is essential for both equations.

We apply tom Dieck's retraction contruction to $Z = \tilde X$ and $z = * \in X_{01} = \tilde X_{01}$ by first chosing $u_x$ in $\Pi(X_{01})$ for all $x \in X_{01}$ (where of course $u_* = id_*$), then $u_x$ in $\Pi(\tilde X_0)$ for all $x \in \tilde X_0 \setminus X_{01}$ and finally $u_x$ in $\Pi(\tilde X_1)$ for all $x \in \tilde X_1 \setminus X_{01}$. Since $\Pi(X_{01}),\Pi(\tilde X_0), \Pi(\tilde X_1)$ are subcategories of $\Pi(\tilde X)$, this gives us a choice of $u_x$ in $\Pi(\tilde X)$ for all $x \in \tilde X$ providing a retraction $\tilde r : \Pi(\tilde X) \to \Pi(\tilde X,*) = \Pi(X,*)$. We extend it to a retraction $r : \Pi(X) \to \Pi(X,*)$ as follows: Given a morphisms $\alpha : x \to y$ in $\Pi(x)$, then $x,y$ belong to same path component $P$ of $X$. If $P = \tilde X$, we define $r(\alpha) = \tilde r(\alpha)$. If $P \ne \tilde X$, we define $r(\alpha) =id_*$. Consider the restriction $r_{01}: \Pi(X_{01}) \to \Pi(X,*)$. The category $\Pi(X_{01})$ is a subcategory of $\Pi(\tilde X,*)$ and by construction $r_{01}(\Pi(X_{01})) = \tilde r(\Pi(X_{01})) \subset \Pi(X_{01},*)$, i.e. we may regard $r_{01}$ as a map $r_{01} : \Pi(X_{01}) \to \Pi(X_{01},*)$. Next consider the restriction $r_0 : \Pi(X_0) \to \Pi(X,*)$. For the subcategory $\Pi(\tilde X_0) \subset \Pi(X_0)$ we have by construction $r_0(\Pi(\tilde X_0)) = \tilde r(\Pi(\tilde X_0)) \subset \Pi(X_0,*)$. Let $\tilde P_0$ be a path component of $X_0$ different from $\tilde X_0$, i.e. $\tilde P_0 \cap \tilde X_0 = \emptyset$. Then $\tilde P_0 \cap \tilde X = \tilde P_0 \cap \tilde X_1 = \tilde P_0 \cap X_0 \cap \tilde X_1 \subset \tilde P_0 \cap X_0 \cap X_1 = \tilde P_0 \cap \tilde X_0 \cap \tilde X_1 \subset \tilde P_0 \cap \tilde X_0 = \emptyset$. Thus $\tilde P_0 \cap \tilde X = \emptyset$ and therefore by construction $r_0(\Pi(\tilde P_0)) = r(\Pi(\tilde P_0) = \{id_*\} \subset \Pi(X_0,*)$. We conclude $r_0(\Pi(X_0)) \subset \Pi(X_0,*)$, i.e. we may regard $r_0$ as a map $r_0 : \Pi(X_0) \to \Pi(X_0,*)$. Similarly $r$ restricts to $r_1 : \Pi(X_1) \to \Pi(X_1,*)$. Therefore we get a commutative diagram

\begin{matrix}\Pi(X_0) & \hookleftarrow & \Pi_1(X_{01}) & \hookrightarrow & \Pi(X_1)\\ \downarrow r_0 && \downarrow r_{01} &&\downarrow r_1 \\ \Pi(X_0,*) & \hookleftarrow & \Pi(X_{01},*) & \hookrightarrow & \Pi(X_1,*) \end{matrix}

Now the same argument as in the proof of tom Dieck's Theorem (2.6.2) applies.

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I think I knew why $\tilde{X} \subset \tilde{X}_0 \cup \tilde{X}_1$.

Let $X_0$ and $X_1$ be subspaces of $X$ such that the interiors cover X. Suppose $X_{01}=X_0\cap X_1$ is path connected.

For any point $x\in \tilde{X}$, there is a path $a:I\to \tilde{X}$ such that $a(0)=\ast$ and $a(1)=x$. We can find $0=t_0<t_1<...<t_{n-1}<t_n=1$ such that for all $i=0,...,n-1$, $a([t_i,t_{i+1}])\subset$int$(X_v)$ for some $v=0,1$. (Lebesgue number of open covering is used here.) We may assume that $a([t_{n-1},t_n])\subset$ int$(X_0)$. There is a minimal integer $k$ such that $a([t_k,t_n])\subset$ int($X_0$). Hence $a(t_k)\in X_{01}$. There is a path $b:I\to X_{01}$ such that $b(0)=\ast$ and $b(1)=a(t_k)$ since $X_{01}$ is path connected. Then the product path of $b$ and $a|{[t_k, t_n]}$ is a path in $X_0$ connecting $\ast$ and $x$. Hence $x\in \tilde{X}_0$. And the path connectedness of $X_{01}$ is crucial here.

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  • $\begingroup$ Yes! This is what I had in mind! $\endgroup$
    – Tyrone
    Apr 21, 2021 at 14:55

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