张柏芝综艺T台秀 精致魅惑彰显自信风采

Question

This looks like a question that could be proved using the Seifert–Van Kampen theorem, but I am not sure if it is true.

Let $X=U \cup V$ be a topological space, where $U$, $V$ and $U\cap V$ are open, path-connected subspaces of $X$. Let $$i_U: \pi_1(U \cap V) \rightarrow \pi_1(U)\quad \text{and}\quad i_V: \pi_1(U \cap V) \rightarrow \pi_1(V)$$ be the inclusion maps. Suppose there exists a homotopically nontrivial element $\alpha \in \pi_1(U \cap V)$ such that $i_U(\alpha)$ is also nontrivial in $\pi_1(U)$. Is it true that if the inclusion map $i_V$ is injective, then $\alpha$ is nontrivial in $\pi_1(X)$?

Edit 1: I tried using the amalgamated product of fundamental groups of $U$ and $V$ to prove this result. Roughly speaking $\pi_1(X) = \langle \pi_1(U), \pi_1(V) : i_V(\alpha_i) = i_U(\alpha_i) \rangle$ where $\alpha_i$ are the generators of the intersection $\pi_1(U\cap V)$. By assumption none of the additional relations in the finite presentation of the fundamental group is of the form $\alpha = 1$. But I find it difficult to prove that the relations in the groups $\pi_1(U)$ and $\pi_1(V)$ cannot be used with the additional relations in the amalgamated product to yield $\alpha =1$.

Edit 2: Following the suggestion in the comments, the group-theoretic version of this question can be found here.

Answer 1

Here is an intuitive counterexample.

Consider two U-shaped curves in $\mathbb{R}^3$:

• $K_1$, which extends upward, and
• $K_2$, which extends downward and hooks around $K_1$,

as shown in the figure. Assume that one end of $K_1$ has finite length, while the other extends infinitely. Let $X$ be the complement of these curves,

$$ X = \mathbb{R}^3 \setminus (K_1 \cup K_2), $$

and define open subsets $U, V \subset X$ by

$$ U = \{(x, y, z) \in X \mid z > 0\}, \quad V = \{(x, y, z) \in X \mid z < 1\}. $$

In this setup, the fundamental groups of $U \cap V$, $U$, and $V$ are as follows:

• $\pi_1(U \cap V) = \langle \alpha, \beta \rangle$.
• $\pi_1(U)=\langle \alpha\rangle$. The natural map $i_U \colon \pi_1(U \cap V) \to \pi_1(U)$ is given by $\alpha \mapsto \alpha$, $\beta \mapsto 1$.
• $\pi_1(V) = \langle \alpha, \gamma\rangle$. The natural map $i_V \colon \pi_1(U \cap V) \to \pi_1(V)$ is given by $\alpha \mapsto \alpha$, $\beta \mapsto \gamma\alpha\gamma^{-1}$, and is therefore injective.

However, from the figure, one can see that $\alpha$ can be contracted within $X$.