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Let $V$ is a projective hypersurface of dimension $3$ and $D$ be divisor at infinity of $V$ (assume $D$ has isolated singularities). It is known that the third homology of both $V$ and $D$ are hard to find, so I assume $H_3(V; Z)=Z$$H_3(V; \mathbb{Z})= \mathbb{ Z}$ and $H_3(D; Z)=Z$$H_3(D; \mathbb{ Z})= \mathbb{ Z}$.

My question is as follows: Is it possible that the inclusion $i: D\to V$ induces an isomorphism in all homology up to $4,$ except the third one is multiplication by $p,$ where $p$ is prime?

Let $V$ is a projective hypersurface of dimension $3$ and $D$ be divisor at infinity of $V$ (assume $D$ has isolated singularities). It is known that the third homology of both $V$ and $D$ are hard to find, so I assume $H_3(V; Z)=Z$ and $H_3(D; Z)=Z$.

My question is as follows: Is it possible that the inclusion $i: D\to V$ induces an isomorphism in all homology up to $4,$ except the third one is multiplication by $p,$ where $p$ is prime?

Let $V$ is a projective hypersurface of dimension $3$ and $D$ be divisor at infinity of $V$ (assume $D$ has isolated singularities). It is known that the third homology of both $V$ and $D$ are hard to find, so I assume $H_3(V; \mathbb{Z})= \mathbb{ Z}$ and $H_3(D; \mathbb{ Z})= \mathbb{ Z}$.

My question is as follows: Is it possible that the inclusion $i: D\to V$ induces an isomorphism in all homology up to $4,$ except the third one is multiplication by $p,$ where $p$ is prime?

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piper1967
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Inclusion inducing isomorphism at all level except one

Let $V$ is a projective hypersurface of dimension $3$ and $D$ be divisor at infinity of $V$ (assume $D$ has isolated singularities). It is known that the third homology of both $V$ and $D$ are hard to find, so I assume $H_3(V; Z)=Z$ and $H_3(D; Z)=Z$.

My question is as follows: Is it possible that the inclusion $i: D\to V$ induces an isomorphism in all homology up to $4,$ except the third one is multiplication by $p,$ where $p$ is prime?