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Consider the definition of CW complex from hatcher I am trying to understand the issue with the identification, because I feel there is something I don't understand. I decided to do an example and do all the detailed calculation. Suppose we want to construct $S^1$ as a CW complex according to hatcher ingridents we start with a 0-cell, so in our case it will just be a single point, so here is our procedures:

1)$X_0 = \{x_0\}$.

2)$X_1 = (X_0 \sqcup \mathbb{D^1})/\backsim$, where the ~ is defined as

$x \backsim \phi(x)$, where $\phi(x) : S^0 \rightarrow X_0$, since $X_0$ has only one point, which is $x_0$ and $S^0 = \{-1,1\}$, so both -1 and 1 will go to $x_0$, so we have $-1 \backsim x_0$ and $1 \backsim x_0$, hence $[x_0] = \{-1,1\}$, but shouldn't we also have $x_0$ in the equivalence class as well ? What about the other elements of $X_1$ that aren't in boundary where are they mapped to ? I guess geometrically what we are doing is that we are attaching the end point of a line to a common point and we wrap around it, however I want to understand of why this is actually the case.

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  • $[x_0]=[1]=[-1]$ because the total space we are quotienting is $X_0 \sqcup \mathbb{D}^1$. Is your question about how the non-boundary points of $\mathbb{D}^1$ behave under the quotient identification? If so, in general we only emphasize those points which are identified for the construction, and for those points we verbally omit we identify them with only themselves, $x \sim y$ if and only if $x=y$ for all $x \in \text{int} \mathbb{D}^1$. – Yeldarbskich Jan 30 '16 at 05:18
  • Alright I understand now thank you. –  Jan 30 '16 at 05:26

1 Answers1

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It seems you're having trouble in the definition of quotient spaces, not in the definition of CW-complexes. If $X$ is a topological space, $\sim$ an equivalence relation on $X$, $X/\sim$ is defined to be the space with underlying set consisting of the $\sim$-equivalence classes of $X$ and topology coming from the final topology on the quotient map $q : X \to X/\sim$.

That said, $S^1 = \{x_0\} \coprod D^1/\sim$ where $\sim$ is the equivalence relation defined by $1 \sim x_0 \sim -1$. So all three points $1, x_0, -1$ in the disjoint union $\{x_0\} \coprod D^1$ are mapped to the same point in $\{x_0\} \coprod D^1/\sim$. Geometrically, this corresponds to gluing $x_0$ and the two endpoints of the interval to a single point and leaving everything as is. Geometrically it is clear why it's a circle.

Balarka Sen
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