On P.138 of Hatcher (P.147 of pdf), Hatcher claims that (paraphrase):

Let $X$ be an infinite-dimensional CW complex. Let $X^n$ denote the $n$-skeleton of $X$, i.e. the cells with $n$ dimensions or below. Identify $X^n$ as a subset of $X^{n+1}$. Let $T = \bigcup_{n \in \Bbb N} X^n \times [n,\infty)$. Then, $X$ and $T$ are homotopy equivalent.

I see a map from $T$ to $X$, but I don't see how one builds a map from $X$ to $T$, let alone showing that they are homotopy-inverses of each other.

In Hatcher's proof, the time $t=1$ of the homotopy is left unspecified, and he never constructed a map from $X$ to $T$.

  • $\begingroup$ There isn't an obvious map in the other direction. Can you show that your map $T\rightarrow X$ is a weak equivalence using the fact that $S^n$ is compact? If you can, then you can appeal to Whitehead's theorem. $\endgroup$ – Tyrone Mar 26 '18 at 17:14

This is the most nonsense I've ever spewed. I apologize to the readers sincerely.

Review the definition of a CW-complex. We will define finite-dimensional CW-complexes first, and then use that to build our infinite-dimensional CW-complexes.

Let $I_n$ be an indexing set for each $n \in \Bbb N$. We build our stages $X^n$ recursively until we reach the dimension of the complex we want.

$X^0$ is defined to be $I_0$ with the discrete topology.

$X^k$ having been assumed to be defined, we require maps $\varphi^k_\alpha : \partial D^{k+1} \to X^k$ with index $\alpha \in I_k$ to advance to the next stage, wherein we define $X^{k+1}$ to the colimit of the following diagram: $$X^k\overset {\varphi^k_\alpha} \longleftarrow \partial D^{k+1} \overset {i^k_\alpha} \longrightarrow D^{k+1}$$ Note that there are as many copies of $\partial D^{k+1}$ and $D^{k+1}$ in the diagram as the indexing set $I_k$, but only one copy of $X^k$.

Being a colimit, $X^{k+1}$ comes with these maps:

  • $iL^k : X^k \to X^{k+1}$
  • $iR^k_\alpha : D^{k+1} \to X^{k+1}$

It has the property that $iL^k \circ \varphi^k_\alpha = iR^k_\alpha \circ i^k_\alpha$.

It also has the property that for every $T$ with maps $tL^k : X^k \to T$ and $tR^k_\alpha : D^{k+1} \to T$ for each $\alpha$ such that $tL^k \circ \varphi^k_\alpha = tR^k_\alpha \circ i^k_\alpha$, there is a unique map $\pi_{k+1}: X^{k+1} \to T$ such that $tL^k = \pi_{k+1} \circ iL^k$ and $tR^k_\alpha = \pi_{k+1} \circ iR^k_\alpha$.

Now, we have defined finite-dimensional CW complexes.

To obtain an infinite-dimensional CW complexes, require that we have $X^n$ for each $n \in \Bbb N$, and take the direct limit of $X^n$ with the maps being $iL^k : X^k \to X^{k+1}$. Call the direct limit $X$, and by universal properties we have maps $iX^k : X^k \to X$, and the property that if we have maps $\pi_k : X^k \to T$ such that $\pi_k = \pi_{k+1} \circ iL^k$, then there is a unique map $\pi : X \to T$ such that $\pi_k = \pi \circ iX^K$.

Now we wish to build $\pi : X \to T$, where $T = \bigcup_{n \in \Bbb N} X^n \times [n,\infty)$ identified as a subset of $X \times [0,\infty)$, with identifications done with $iL^k$.

Before we proceed, I will build some auxiliary functions $\psi_n : X^n \times [0, n] \to T \cap (X^n \times [0, n])$ by recursion.

  1. $\psi_0$ is the identity function.

  2. Given $\psi_k$, we construct $\psi_{k+1} : X^{k+1} \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$ as follows: By the property of $X^{k+1}$, and that colimit commutes with product if the multiplicand ($[0, k+1]$) is locally compact (reference), we need to construct $tL^k : X^k \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$ and $tR^k_\alpha : D^{k+1} \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$ for each $\alpha$ such that $tL^k \circ (\varphi^k_\alpha \times \operatorname{id}_{[0, k+1]}) = tR^k_\alpha \circ (i^k_\alpha \times \operatorname{id}_{[0, k+1]})$.

  3. To construct $tL^k : X^k \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$, we let it to be $\psi_k$ on $X^k \times [0, k]$ and $\operatorname{id}$ on $X^k \times [k, k+1]$. We will leave an asterisk here and show later that this is well-defined.

  4. To construct $tR^k_\alpha : D^{k+1} \times [0, k+1] \to T \cap (X^{k+1} \times [0, k+1])$, we let $(\vec v, (k+1) s) \in D^{k+1} \times [0, k+1]$. If $\|\vec v\| \le \frac {1 + s} 2$, then send it to $(iR^k_\alpha \left( \frac {2v} {1+s} \right), k+1)$. If $\|\vec v\| \ge \frac {1 + s} 2$, then send it to $tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), (k+1)(s+2(1-\|\vec v\|)))$. Well-defined: when $\| \vec v \| = \frac {1+s} 2$: $$ \begin{array}{cl} & tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), (k+1)(s+2(1-\|\vec v\|))) \\ =& tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\ =& (\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\ =& (iL^k (\varphi^k_\alpha(\vec v/\|\vec v\|)), k+1) \\ =& (iR^k_\alpha (i^k_\alpha(\vec v/\|\vec v\|)), k+1) \\ =& (iR^k_\alpha (\vec v/\|\vec v\|), k+1) \\ =& (iR^k_\alpha \left( \frac {2v} {1+s} \right), k+1) \end{array}$$

  5. To resolve the asterisk in step 3, i.e. to prove that $\psi_n(x, n) = (x, n)$ for $x \in X^n$, we use induction (and clear that asterisk by using the induction hypothesis). Now, it is true for $n=0$ since $\psi_0$ is the identity function. For $n=k+1$, let $x \in X^{k+1}$. If $x = iL^k(x')$: $$\begin{array}{cl} & \psi_{k+1}(x, k+1) \\ =& \psi_{k+1}(iL^k(x'), k+1) \\ =& tL^k(x', k+1) \\ =& (x', k+1) \\ =& (iL(x'), k+1) \\ =& (x, k+1) \end{array}$$ If $x = iR^k_\alpha(\vec v)$, note that the casing parameter $\frac{1+s}2$ is equal to $1$ when $s=k+1$, so the first case in $tR^k$ always matches: $$\begin{array}{cl} & \psi_{k+1}(x, k+1) \\ =& \psi_{k+1}(iR^k_\alpha(\vec v), k+1) \\ =& tR^k_\alpha(\vec v, k+1) \\ =& (iR^k_\alpha(\vec v), k+1) \\ =& (x, k+1) \end{array}$$

Now we can finally build $\pi : X \to T$.

  1. By property of $X$, we need $\pi_n : X^n \to T$ such that $\pi_n = \pi_{n+1} \circ iL^n$.

  2. Build $\pi_n$ by induction.

  3. Build $\pi_0$ by sending $x \in X^0$ to $(x,0)$.

  4. We build $\pi_{k+1} : X^{k+1} \to T$ by letting $\pi_{k+1}(x) := \psi_{k+1}(x, 0)$.

  5. To check that $\pi_n = \pi_{n+1} \circ iL^n$, let $x \in X^n$: $$\begin{array}{cl} & \pi_{n+1}(iL^n(x)) \\ =& \psi_{n+1}(iL^n(x), 0) \\ =& tL^k(x, 0) \\ =& \psi_k(x, 0) \\ =& \pi_k(x) \\ \end{array}$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.