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I am trying to understand a bit the concept of CW complexes. It seems that simplicial commplexes are cell complexes as well. I would have said intuitively that in that case, the cells $e_{\alpha}$ are the interior of the simplices $Int (\sigma)$ of the simplicial complexes. But in Munkres' book I understand that that the cells $e_\alpha$ are union of open simplices. Which one is correct?


We now see how strong the analogy is between the homology of simplicial complexes and the homology of CW complexes. Let use introduce some terminology that will make the analogy even stronger. For each open p-cell $e_{\alpha}$ of the CW complex X, the group $H_p(\bar{e_{\alpha}}, \dot{e_{\alpha}})$ is infinite cyclic. The two generators of this group will be called the two orientations of $e_{\alpha}$. An oriented p-cell of X is an open p-cell $e_{\alpha}$ together with an orientation of $e_{\alpha}$. The cellular chain group $D_p(X)=H_p(X^p,X^{p-1})$ is a free abelian group. One obtains a basis for it by orienting each open p-cell $e_{\alpha}$ of $X$ and passing to the corresponding element of $H_p(X^p,X^{p-1})$. [That is, by taking the image of the orientation under the homomorphism induced by inclusion $H_p(\bar{e_{\alpha}}, \dot{e_{\alpha}}) \to > H_p(X^p,X^{p-1}).$]

The homology of the chain complex $D(X)$ is isomorphic, by our theorem, with the singular homology of X. In the special case where $X$ is a triangulable CW-complex triangulated by a complex $K$, and $H_p$ denotes simplicial homology, we interpret these comments as follows : The fact that $X^p$ and $X^{p-1}$ are subcomplexes of $K$ implies that each open p-cell $e_{\alpha}$ is a union of open simplices of $K$, so that $\bar{e_{\alpha}}$ is the polytope of a subcomplex of $K$. The group $H_p(\bar{e_{\alpha}}, \dot{e_{\alpha}})$ equals the group of p-chains carried by $\dot{e_{\alpha}}$. This group is infinite cyclic; either generator of this group is called a fundamental cycle for $(\bar{e_{\alpha}}, \dot{e_{\alpha}})$.

Why the cells are union of open simplices and not just open simplices? I don't really get what the fundamental cycles are.

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  • $\begingroup$ I thought a simplex was just a specific choice of cell type. You could also choose cubical cells, for example. $\endgroup$
    – rschwieb
    Commented Jul 1, 2019 at 16:28
  • $\begingroup$ are union of open simplices This sounds, somewhat, like a simplicial complex. I think the individual simplices are also required to satisfy additional requirements on their position, e.g. that their intersections are simplices or are empty. i'm an amateur at homology though, so I could be wrong. $\endgroup$
    – rschwieb
    Commented Jul 1, 2019 at 16:33
  • $\begingroup$ It would help if you provided the exact quote from Munkres. $\endgroup$
    – Lee Mosher
    Commented Jul 1, 2019 at 21:12
  • $\begingroup$ I don't see anywhere in this quote where it says that "the cells $e_\alpha$ are union of open simplices". $\endgroup$
    – Lee Mosher
    Commented Jul 3, 2019 at 16:32
  • $\begingroup$ Thanks, I edited $\endgroup$
    – roi_saumon
    Commented Jul 3, 2019 at 16:36

1 Answer 1

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Munkres considers a triangulable CW-complex $X$. This means that you have

  1. a CW-structure $\mathfrak C$ on $X$ with open cells $e_\alpha^n$ (where $n$ indicates the dimension of the cell)

  2. a triangulation $K$ of $X$ with (compact) simplices $\sigma_\beta^n$

The open simplices $s_\beta^n \subset\sigma_\beta^n$ give us a CW-structure $\mathfrak C'$ on $X$, but there is no reason why $\mathfrak C = \mathfrak C'$. As an example consider $S^1 \subset \mathbb R^2$ with the CW-structure $\mathfrak C$ given by the two open cells $p =\{(0,1)\},S^1 \setminus \{p\}$ and the triangulation $K$ given by the six simplices $$p, q =\{(1,0)\},-q, A = \{(x,y) \in S^1 \mid x \ge 0, y \ge 0 \}, B = \{(x,y) \in S^1 \mid x \ge 0, y \le 0 \}, \\C = \{(x,y) \in S^1 \mid x \le 0 \} .$$

Now it seems that Munkres considers triangulations of a special kind which is indicated by "The fact that $X^p$ and $X^{p−1}$ are subcomplexes of $K$ ..." As an example that this is not always true consider again $S^1$ with $\mathfrak C$ and the triangulation $L$ given by the six simplices $$-p, q,-q, D = \{(x,y) \in S^1 \mid x \le 0, y \ge 0 \}, E= \{(x,y) \in S^1 \mid x \le 0, y \le 0 \}, \\F = \{(x,y) \in S^1 \mid x \ge 0 \} .$$ Then the $0$-skeleton of $S^1$ is not a subcomplex of $L$.

However, under Munkres' assumption we can easily verify that each open $p$-cell of $X$ is a union of open simplices of $K$ (but in general not an open simplex of $K$ as shown by the above example).

If each $X^p$ is triangulated by a subcomplex $K^p$ of $K$, then $X^p$ is the union of all open simplices of $K^p$ and $X^{p-1}$ is the union of all open simplices of $K^{p-1}$. Hence $X^p \setminus X^{p-1}$ is the union of all open simplices $s_\beta$ of $K^p$ which do not belong to $K^{p-1}$. Now $X^p \setminus X^{p-1}$ is the union of the open $p$-cells $e^p_\alpha$ of $X$ which are pairwise disjoint and connected. Since each $s_\beta$ is connected, it is contained in a unique $e^p_\alpha$. This proves the above claim.

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