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Let $M_1$ and $M_2$ be two compact manifolds of dimension $n\ge 3$. Let us have embeddings $i_1: K \to M_1$ and $i_2: K \to M_2$ for a closed manifold $K$ of dimension at most $n-1$ such that the normal bundles $N_1$ and $N_2$ corresponding to $i_1$ and $i_2$, respectively, are isomorphic. Let $U_1$ and $U_2$ be tubular neighborhoods of $i_1(K)$ and $i_2(K)$, that are identified with the total spaces of $N_1$ and $N_2$, respectively. It is easy to find a homeomorphism $f: U_1 - i_1(K) \to U_2 - i_2(K)$. Let us obtain $M_1\#_K M_2$ from $(M_1 - i_1(K))\cup(M_2 - i_2(K))$ by gluing $U_1 - i_1(K)$ and $U_2 - i_2(K)$ along $f$. This is the connected sum of $M_1$ and $M_2$ along the submanifold $K$ described here. Let $M = M_1 \cup_K M_2$ be the union of $M_1$ and $M_2$ along $K$.

I have the following related questions about this construction.

  1. Is $M_1 \#_K M_2$ always a closed (compact with boundary) manifold? If not in general, then does it at least hold if both $M_1$ and $M_2$ are closed?
  2. Is the quotient $M/K$ homeomorphic to the wedge $M_1/(U_1 - i_1(K)) \vee M_2/(U_2 - i_2(K))$?
  3. If the answer to the above question is YES, then is the quotient map $g :M \to M/K$ a homotopy equivalence?

All these questions are motivated by the corresponding questions (having positive answers) to the usual connected sum along a point.

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  • $\begingroup$ I don't understand question 2: there's no K in M. $\endgroup$ Commented Sep 1 at 7:40
  • $\begingroup$ @MarcoGolla $K$ is embedded in $M_1$ and $M_2$, and $M$ is the union of $M_1$ and $M_2$ along $K$. Isn't there a copy of $K$ in $M$ then? $\endgroup$
    – Katrina
    Commented Sep 1 at 19:50
  • $\begingroup$ No, you're exactly gluing $M_1\setminus K_1$ and $M_2\setminus K_2$, so there's no $K$. $\endgroup$ Commented Sep 1 at 20:09
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    $\begingroup$ You should ask him, not me :) $\endgroup$ Commented Sep 4 at 7:45
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    $\begingroup$ The natural submanifold of your $M_1 \#_K M_2$ is the sphere bundle $S_K = \partial U_i$ over $K$. One has $(M_1 \#_K M_2)/S_K \cong (M_1/K) \vee (M_2/K)$. Marco is commenting about this setting. You also look at the (non-manifold) $M$ which is simply $M_1 \cup_K M_2$, for which one also has $(M_1 \cup_K M_2)/K \cong (M_1/K) \vee (M_2/K)$. When the normal bundle to $K$ has a section, then $S_K$ contains (perhaps some number of distinct) submanifolds diffeomorphic to $K$. You could look into collapsing these, but that's probably not what you want. $\endgroup$
    – mme
    Commented Sep 10 at 13:39

1 Answer 1

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Note that "closed" means "compact without boundary". Here are answers. (1): Yes, if $M_1$ and $M_2$ are closed manifolds then $M_1\sharp_KM_2$ is a closed manifold. If $M_1$ and $M_2$ are compact manifolds with boundary and if $i_1(K)$ is interior to $M_1$ (i.e. disjoint from the boundary $\partial M_1$) and if $i_2(K)$ is interior to $M_2$, then $M=M_1\sharp_KM_2$ is a compact manifold with boundary. You can prove that by giving an atlas. (2) Your questions (2) and (3) make sense if there is an embedding of $K$ into $M$. This will be guaranteed if the common normal bundle to $i_1$ and $i_2$ admits a nonvanishing section. Then, the answer to (2) is Yes. (3) Probably almost always no, if $K$ is not reduced to one point. For example, still under the hypothesis that the common normal bundle to $i_1$ and $i_2$ admits a nonvanishing section, then you get an inclusion $K\subset M$ as a submanifold which will in general not be homotopic to a constant map; while the mapping $K\to M/K$ is homotopic to a constant map. You can also think to the case $K=$ two points and $M_1=M_2=S^1$.

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  • $\begingroup$ Thank you for your answer. This is in response to a comment by Marco Golla on my question that $M/K$ does not make sense since there is no $K$ in $M$. In your answer, you found part (2) of my question ok and you said that $M/K$ is indeed homeomorphic to the space I described. So, does part (2) of my question still make sense to you? $\endgroup$
    – Katrina
    Commented Sep 4 at 13:56

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