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Studying for quals and came across this question:

Let $X_n$ be formed by taking disjoint unions of $n$ cylinders ($S^1$ x $I$) say $C_1, C_2, ... C_n$, by gluing for each $k$, the $S^1$ x $\{1\}$ of $C_k$ to the $S^1$ x $\{0\}$ of $C_{k+1}$ along a map of degree $k$. Taking $X$ to the direct limit (under inclusions), compute $H_1(X)$.

$\textbf{Attempted Solution:}$ $X_1$, $X_2$ are just usual cylinders and then using Mayer-Vietoris for each subsequent $X_n$ can get $$H_1(X_n) = \mathbb{Z} \oplus \bigoplus_{i = 2}^{n-1}\mathbb{Z}_i $$

Then for $X$, we just get the same but with the summation for $i\geq 2$.

Just wanted to confirm if this is correct (or am I missing some subtlety about the direct limit).

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    $\begingroup$ Note that a mapping cylinder of $X \to Y$ deformation retracts to $Y$, so by induction each $X_n \simeq S^1$ $\endgroup$
    – ronno
    Commented May 13 at 6:50

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As ronno points out in the comments, what you wrote is not correct: each $X_n$ deformation retracts onto $S^1$, so $H_1(X_n) \cong \mathbb{Z}$.

The way mapping telescope computations like this work is to then realize that the induced map $H_1(X_n) \to H_1(X_{n + 1})$ is multiplication by $n$ (this you should try and prove yourself if it's not immediately clear to you), so $$ \varinjlim_n H_1(X_n) \cong \varinjlim (\mathbb{Z} \overset{1}{\to} \mathbb{Z} \overset{2}{\to}\mathbb{Z} \overset{3}{\to} \cdots) \cong \mathbb{Q} $$ and the last thing you need to do is to pull the limit into $H_1$, which you hopefully know you're allowed to do (see e.g. Proposition 3.33 in Hatcher).

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    $\begingroup$ Totally forgot about the deformation retraction. As for the multiplication, I understand it's given because of the degree $n$ mapping between the circles. Thanks! $\endgroup$ Commented May 13 at 18:37

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