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Let $Y$ be a rational homology three sphere, and $Y\hookrightarrow S^1\times S^3$ a smooth embedding.

Q. Can we say that $[Y]$ is a generator of $H_3(S^1\times S^3;\mathbb Z)$?

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    $\begingroup$ Of course not. Perhaps you mean to ask a more refined question? $\endgroup$ – Ryan Budney Dec 17 '17 at 18:27
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If it's nonseparating, then it is a generator. This is a straightforward consequence of Poincare duality. If it separates then of course it is 0 in homology.

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Suppose $Y$ is the standard three sphere. Since $S^1\times S^3$ is a four dimensional manifold, it contains an embedded $\mathbb{R}^4$ (a chart). Just embed $Y$ into this $\mathbb{R}^4$.

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