# Are the path connectedness of $X_0$ and $X_1$ necessary in Seifert-van Kampen theorem?

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 ?

• 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. Apr 20, 2021 at 14:36
• @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? Apr 20, 2021 at 14:45
• 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 $X_0,X_1$. Apr 20, 2021 at 15:03
• @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? Apr 21, 2021 at 0:03
• 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). Apr 22, 2021 at 13:53

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.

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 $$0=t_0 such that for all $$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.

• Yes! This is what I had in mind! Apr 21, 2021 at 14:55