Let $(X, \Gamma)$ be a cell complex, and let $e \in \Gamma$ be a open cell of dimension $n \geq 1$.

I want to try and construct an example of a characteristic map $\Phi : D \to X$ (where $D$ is some closed $n$-cell) for which $\Phi[D] = \operatorname{Cl}_X(e)$ is not a closed cell. To do that I need to find a $\Phi$ that is not injective on the boundary of $D$.

I am not sure how to construct such an example (and in this example, I'd need to choose, $X$, $\Gamma$, $e$ and $\Phi$ appropriately so that $\Phi$ is not injective on the boundary)

The biggest problem is having that $\Phi$ must be continuous, for example we could take $X = [0, 1]$ and $\Gamma = \{(0, 1) \{0\}, \{1\}\}$ and define $\Phi : [0, 1] \to [0, 1]$ by $\Phi(x) = x$ for $x \in (0, 1)$ and $\Phi(x) = 0$ on $x \in \{0, 1\}$, and $\Phi$ satisfies all the criteria for being a characteristic map except for being continuous.

Note that this is not a homework problem, but rather a problem that I came up with while reading a claim in a book and wanting to find an explicit example for myself

  • $\begingroup$ What's the usual cell structure on the circle? $\endgroup$ – Steve D Feb 26 '18 at 18:18

Here's the simplest example. Let $X$ be the circle $\{z\in\mathbb{C}:|z|=1\}$ and put a cell complex structure on it as follows. We have one $0$-cell, which is just $\{1\}$. We have one $1$-cell, whose characteristic map $\Phi:[0,1]\to X$ is defined by $\Phi(t)=e^{2\pi i t}$. This is not injective, since $\Phi(0)=\Phi(1)$.

  • $\begingroup$ Thanks for the answer! Just a quick question. don't you mean $z \in \mathbb{C}$ in your answer? $\endgroup$ – Perturbative Feb 26 '18 at 18:26
  • $\begingroup$ Oops, yes, I've fixed that. $\endgroup$ – Eric Wofsey Feb 26 '18 at 18:54

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