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change some error in the mathematical state and words.

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edit approved Oct 31, 2021 at 11:22

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 $$\pi_1(X_0) \longrightarrow \pi_1(X\setminus K) \longrightarrow \pi_1(X)$$

ToIn 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 geodesically completea 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 $X$.

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 homomorphisms $$\pi_1(X_0) \longrightarrow \pi_1(X\setminus K) \longrightarrow \pi_1(X)$$

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. Therefore $X_0$ is the union of all the geodesics in $X_0$ starting from $p$. Since $X_0$ is geodesically complete, 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 $X$.

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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created Oct 19, 2021 at 12:10

Bruno Martelli

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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 homomorphisms $$\pi_1(X_0) \longrightarrow \pi_1(X\setminus K) \longrightarrow \pi_1(X)$$

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. Therefore $X_0$ is the union of all the geodesics in $X_0$ starting from $p$. Since $X_0$ is geodesically complete, 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 $X$.