# Mapping cone, Homology groups and universal cover

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

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

I know that the mapping cone $C_f$ of $f: X \rightarrow Y$, is defined to be the quotient of the mapping cylinder of $f$ with $X$.
Or we can say,
Given a map $f: X \rightarrow Y$, the mapping cone $C_f$ is defined to be the quotient of the topological space of $(X \times I) \coprod Y$ with respect to the equivalence relation $(x,0) \sim (x',0), (x,1) \sim f(x)$ , on $X$. Here $I$ denotes the unit interval $[0,1]$ 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 $W$ and of its universal cover.
• Dec 5th 2010, 06:58 PM
Tinyboss
The "cone" part is contractible to the "point" $X\times\{0\}$, 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 $W$ deformation retracts to $S^1 \times \{0 \}$, so they are homotopic equivalent, hence $H_n(W) = H_n(S^1 \times \{0 \} ) = Z$ if n=1 or 0 and 0 if n>1 ?

What would be the universal cover of $W$?