# Homology of mapping telescope calculation

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).

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

## 1 Answer

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).

• 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! Commented May 13 at 18:37