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Let $M$ ,$N$ be $n$-dimensional manifolds. Let $D_{1},D_{2},\dots D_{k} $ be $n$-dimensional manifolds embedded in $M$ and $\cup_{i=1}^kD_{i}=M$ and each $D_i$ is homeomorphic to $N$. Question is following.

Problem Consider all embedding of $\cup_{i=1}^kD_{i}$. Determine the minimum of the value $k$.

Example When we consider $M$ is 2-dimensional torus and each $D_i$ is a 2-dimensional disk, the minimum value is 3. I thought $M=L(5,2)$ and each $D_i$ is a 3-dimensional ball, and I couldn't determine the minimum.

Can you answer?

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A first observation is that such a $k$ may not exist, for example if $N$ does not embed in $M$.

When $N$ is a disk, then $k$ equals the ball category of $M$, denoted $\operatorname{ballcat}(M)$ or $\operatorname{bcat}(M)$ in the literature. See for instance

Gavrila, Caius, The Lusternik-Schnirelmann theorem for the ball category, Cornea, O. (ed.) et al., Lusternik-Schnirelmann category and related topics. Proceedings of the 2001 AMS-IMS-SIAM joint summer research conference on Lusternik-Schnirelmann category in the new millennium, South Hadley, MA, USA, July 29-August 2, 2001. Providence, RI: American Mathematical Society (AMS). Contemp. Math. 316, 113-119 (2002). ZBL1029.58007

or Chapter 3 of the book Lusternik-Schnirelmann category by Cornea, Lupton, Oprea and Tanré.

In particular if $M$ is a smooth manifold then there is a chain of inequalities $$ \operatorname{Crit}(M)\geq \operatorname{ballcat}(M)\geq \operatorname{cat}(M), $$ where $\operatorname{cat}(M)$ is the Lusternik-Schnirelmann category of $M$ (the minimal $k$ such that $M$ has a cover by $k$ open sets which are contractible in $M$) and $\operatorname{Crit}(M)$ denotes the minimal number of critical points for any smooth function on $M$.

In your example of $M=L(5,2)$ a $3$-dimensional lens space, $\operatorname{cat}(M)=4$ by a result of Fadell and Husseini and $\operatorname{Crit}(M)\le 4$ by a result of Takens (in general for a connected $n$ manifold $\operatorname{Crit}(M)\le n+1$). Hence $\operatorname{ballcat}(M)=4$.

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    $\begingroup$ If $M$ is connected and closed, and there exists an embedding $N\rightarrow M$, then the number is finite right? $\endgroup$
    – Thomas Rot
    Commented Jun 5, 2019 at 9:46
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    $\begingroup$ @ThomasRot: I guess so, since we ought to be able to find embeddings $N_i\hookrightarrow M$ whose interiors cover $M$, then use compactness. $\endgroup$
    – Mark Grant
    Commented Jun 5, 2019 at 10:01
  • $\begingroup$ ballcat(M)=4 is interesting result for me. thank you. Ball category is seem to have application for the number of surgeries to get desired manifold. I will consult it. $\endgroup$ Commented Jun 8, 2019 at 9:25

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