# Mapping cone, Homology groups and universal cover

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• Dec 5th 2010, 10:37 AM
math8
Mapping cone, Homology groups and universal cover
Let $\displaystyle W$ be the mapping cone of the map $\displaystyle f: S^1 \rightarrow S^1$ defined by $\displaystyle f(z)= z^p$.

How do you compute the homology groups of $\displaystyle W$? What about the homology groups of the Universal covering of $\displaystyle W$?

I know that the mapping cone $\displaystyle C_f$ of $\displaystyle f: X \rightarrow Y$, is defined to be the quotient of the mapping cylinder of $\displaystyle f$ with $\displaystyle X$.
Or we can say,
Given a map $\displaystyle f: X \rightarrow Y$, the mapping cone $\displaystyle C_f$ is defined to be the quotient of the topological space of $\displaystyle (X \times I) \coprod Y$ with respect to the equivalence relation $\displaystyle (x,0) \sim (x',0), (x,1) \sim f(x)$ , on $\displaystyle X$. Here $\displaystyle I$ denotes the unit interval $\displaystyle [0,1]$ with its standard topology.

But I am not sure how to start using this definition of the mapping cone, to find the homology groups of $\displaystyle W$ and of its universal cover.
• Dec 5th 2010, 06:58 PM
Tinyboss
The "cone" part is contractible to the "point" $\displaystyle X\times\{0\}$, so its homology is easy to compute. I'd guess that the Mayer-Vietoris sequence might be the way to proceed, but my algebraic topology class just ended and we didn't do mapping cones.
• Dec 6th 2010, 07:09 AM
math8
So, can we say $\displaystyle W$ deformation retracts to $\displaystyle S^1 \times \{0 \}$, so they are homotopic equivalent, hence $\displaystyle H_n(W) = H_n(S^1 \times \{0 \} ) = Z$ if n=1 or 0 and 0 if n>1 ?

What would be the universal cover of $\displaystyle W$?