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Let $M$ be a manifold space, Let $H^1(M,\mathbb R)=\{0\}$, then is it possible to get a submanifold $S$ of $M$. such that $H^1(S, \mathbb R)\neq \{0\}$.

If $M$ is simply connected then we can remove a suitable point and then remaining open subset will be sub manifold and which will have abelian fundamental group and above statement is valid. But for general manifold where $H^1(M)=0$ is not because of simply conectedness, Can we get sub manifold?

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If $U$ is a coordinate neighborhood of $M$, then $U$ is diffeomorphic to an open set in $\Bbb{R}^{\dim M}$. Any such open set will contain a closed loop unless $\dim M = 1$. So, unless $\dim M=1$, $U$ must contain a submanifold $S$ with $H^1(S)=\Bbb{Z}$. (If $\dim M=1$ and $H^1(M)=0$, then $M \cong \Bbb{R}$, which has no such submanifold.)

In general, you can't say much about the homology of $S$ just by knowing something about the homology of $M$; as the example I just gave suggests, there could be arbitrarily weird stuff happening locally.

(Oh, also: if $M$ is simply connected, you can't conclude that $M$ with a point removed will have non-trivial $H^1$. For example, you could take $M=\Bbb{R}^3$.)

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If $M$ is an $n$-manifold then pick a neighborhood homeomorphic to $\mathbb R^n$. This will have $S^1$ as a submanifold, which has $H^1 \ne 0$.