Homotopy equivalence of mapping telescope of CW complex 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 $T$.
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.


$\psi_0$ is the identity function.
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 $tL^k \circ (\varphi^k_\alpha \times \operatorname{id}_{[0, k+1]}) = tR^k_\alpha \circ (i^k_\alpha \times \operatorname{id}_{[0, k+1]})$.
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.
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 $\| \vec v \| = \frac {1+s} 2$:
$$ \begin{array}{cl}
&amp; tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), (k+1)(s+2(1-\|\vec v\|))) \\
=&amp; tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\
=&amp; (\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\
=&amp; (iL^k (\varphi^k_\alpha(\vec v/\|\vec v\|)), k+1) \\
=&amp; (iR^k_\alpha (i^k_\alpha(\vec v/\|\vec v\|)), k+1) \\
=&amp; (iR^k_\alpha (\vec v/\|\vec v\|), k+1) \\
=&amp; (iR^k_\alpha \left( \frac {2v} {1+s} \right), k+1)
\end{array}$$
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 $x = iL^k(x')$:
$$\begin{array}{cl}
&amp; \psi_{k+1}(x, k+1) \\
=&amp; \psi_{k+1}(iL^k(x'), k+1) \\
=&amp; tL^k(x', k+1) \\
=&amp; (x', k+1) \\
=&amp; (iL(x'), k+1) \\
=&amp; (x, k+1)
\end{array}$$
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:
$$\begin{array}{cl}
&amp; \psi_{k+1}(x, k+1) \\
=&amp; \psi_{k+1}(iR^k_\alpha(\vec v), k+1) \\
=&amp; tR^k_\alpha(\vec v, k+1) \\
=&amp; (iR^k_\alpha(\vec v), k+1) \\
=&amp; (x, k+1)
\end{array}$$




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


By property of $X$, we need $\pi_n : X^n \to T$ such that $\pi_n = \pi_{n+1} \circ iL^n$.
Build $\pi_n$ by induction.
Build $\pi_0$ by sending $x \in X^0$ to $(x,0)$.
We build $\pi_{k+1} : X^{k+1} \to T$ by letting $\pi_{k+1}(x) := \psi_{k+1}(x, 0)$.
To check that $\pi_n = \pi_{n+1} \circ iL^n$, let $x \in X^n$:
$$\begin{array}{cl}
&amp; \pi_{n+1}(iL^n(x)) \\
=&amp; \psi_{n+1}(iL^n(x), 0) \\
=&amp; tL^k(x, 0) \\
=&amp; \psi_k(x, 0) \\
=&amp; \pi_k(x) \\
\end{array}$$
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.


$\psi_0$ is the identity function.
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 $tL^k \circ (\varphi^k_\alpha \times \operatorname{id}_{[0, k+1]}) = tR^k_\alpha \circ (i^k_\alpha \times \operatorname{id}_{[0, k+1]})$.
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.
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 $\| \vec v \| = \frac {1+s} 2$:
$$ \begin{array}{cl}
&amp; tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), (k+1)(s+2(1-\|\vec v\|))) \\
=&amp; tL^k(\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\
=&amp; (\varphi^k_\alpha(\vec v/\|\vec v\|), k+1) \\
=&amp; (iL^k (\varphi^k_\alpha(\vec v/\|\vec v\|)), k+1) \\
=&amp; (iR^k_\alpha (i^k_\alpha(\vec v/\|\vec v\|)), k+1) \\
=&amp; (iR^k_\alpha (\vec v/\|\vec v\|), k+1) \\
=&amp; (iR^k_\alpha \left( \frac {2v} {1+s} \right), k+1)
\end{array}$$
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 $x = iL^k(x')$:
$$\begin{array}{cl}
&amp; \psi_{k+1}(x, k+1) \\
=&amp; \psi_{k+1}(iL^k(x'), k+1) \\
=&amp; tL^k(x', k+1) \\
=&amp; (x', k+1) \\
=&amp; (iL(x'), k+1) \\
=&amp; (x, k+1)
\end{array}$$
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:
$$\begin{array}{cl}
&amp; \psi_{k+1}(x, k+1) \\
=&amp; \psi_{k+1}(iR^k_\alpha(\vec v), k+1) \\
=&amp; tR^k_\alpha(\vec v, k+1) \\
=&amp; (iR^k_\alpha(\vec v), k+1) \\
=&amp; (x, k+1)
\end{array}$$




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


By property of $X$, we need $\pi_n : X^n \to T$ such that $\pi_n = \pi_{n+1} \circ iL^n$.
Build $\pi_n$ by induction.
Build $\pi_0$ by sending $x \in X^0$ to $(x,0)$.
We build $\pi_{k+1} : X^{k+1} \to T$ by letting $\pi_{k+1}(x) := \psi_{k+1}(x, 0)$.
To check that $\pi_n = \pi_{n+1} \circ iL^n$, let $x \in X^n$:
$$\begin{array}{cl}
&amp; \pi_{n+1}(iL^n(x)) \\
=&amp; \psi_{n+1}(iL^n(x), 0) \\
=&amp; tL^k(x, 0) \\
=&amp; \psi_k(x, 0) \\
=&amp; \pi_k(x) \\
\end{array}$$