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Let $X$ be a compact manifold of non-positive sectional curvature which carries a connected totally geodesic hypersurface $X_0\subset X$. Let $K$ be any compact subset of $X-X_0$. That's to say we have $K\cap X_0=\emptyset$.

How can I prove that the induced homomorphism between fundamental groups $\pi_1 X_0\to \pi_1(X-K)$ is injective?

I have no idea about it. Could you please give me some help with the detail? thank you very much!

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You need $X_0$ to be closed inside $X$, otherwise, the theorem is false (there are simple counterexamples). So let us suppose that $X_0$ is closed.

Inclusions give the induced homomorphisms $$\pi_1(X_0) \longrightarrow \pi_1(X\setminus K) \longrightarrow \pi_1(X)$$

In order to prove that the first homomorphism is injective, it suffices to prove that the composition $\pi_1(X_0) \to \pi_1(X)$ is injective. So you can forget about $K$.

To prove that $\pi_1(X_0)\to \pi_1(X)$ is injective, consider a point $p\in X_0$ and the exponential map $$\exp_p\colon T_p X \longrightarrow X.$$ Since $X$ has non-positive curvature, this map is a covering (this is the Cartan-Hadamard Theorem). The hypersurface $X_0$ is closed, hence complete, hence geodesically complete by Hopf-Rinow Theorem. Therefore $X_0$ is the union of all the geodesics in $X_0$ starting from the point $p$. Since $X_0$ is a totally geodesic submanifold of $X$, the geodesics in $X_0$ are simply the geodesics in $X$, so finally we conclude that $$X_0 = \exp_p(W)$$ where $W = T_pX_0$ is a vector hyperplane in $T_pX$. Since $W$ is simply connected and $\exp_p$ is the universal covering of $X$, one deduces that $X_0$ is $\pi_1$-injective in $X$.

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    $\begingroup$ Maybe it's helpful to make the following detail explicit. Consider connected topological spaces $Y\subset X$, a universal covering $\widetilde{X}$ for $X$, and a component $\widetilde{Y}$ of the preimage of $Y$ in $X$. The inclusion of $Y$ into $X$ is injective on $\pi_1$ if (and only if) $\widetilde{Y}$ is simply connected. The answer shows that $\widetilde{Y}$ is just the tangent space $W$ to $Y$, so the result follows. $\endgroup$
    – HJRW
    Commented Oct 19, 2021 at 13:25
  • $\begingroup$ What is the counterexample? Could you please state it? $\endgroup$
    – Radeha Longa
    Commented Oct 19, 2021 at 13:49
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    $\begingroup$ I guess $\mathbb{R}^2\setminus\{0\}\subset \mathbb{R}^3$ is a counterexample without closed? $\endgroup$
    – Otis Chodosh
    Commented Oct 19, 2021 at 15:43

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