submanifold with same homology

Question

Suppose $M$ is a manifold without boundary, and $N\subseteq M$ is any submanifold, possibly with boundary. If $H_*(N)\cong H_*(M)$, is it necessarily true that $N\cong M$?

Answer 1

I assume that you want to know whether $M,N$ are homotopically equivalent or not.

First consider a $\epsilon$-neighbourhood of $S^1\vee S^1\vee S^2$ inside $\mathbb R^3$ as our manifold $M$. And consider $N= T^2 $(torus) inside $M$. ( we can do so since $M$ is $3-$dim manifold and torus can be embedded in $\mathbb R^3$). So $H_*(M)=H_*(N)$ but they are not homotopically equivalent. (WHY?)

Now as Mariano suggest what can we say if we know $M$ is closed?

In that case if $M\neq N$ then $N$ is deformation retract onto at least a $n-1$-dim CW-complex. In that case $M,N$ cannot be homotopically equivalent, becasue $H_n(M,\mathbb Z_2)= \mathbb Z_2$ but $H_n(N,\mathbb Z_2)=0$.

Answer 2

No, you can take a point in $R^n$.