Using Serre Spectal sequence to compute $H^*(F)$ where $F$ is homotopy fibre of degree map

Question

As stated in the title I am trying to compute the cohomology ring of $F$, where $F$ is the homtopy fibre of the map $S^k\to S^k$ of degree $n$. I already computed the values of $H^i(F)$ using the Serre Spectral sequence. If I am correct it should come out to: \begin{align*} H^i(F)=\begin{cases} \mathbb{Z} & i=0\\ \mathbb{Z}/n & \text{if }i\text{ multiple of }k\\ 0 & \text{else} \end{cases} \end{align*} Since I used the Serre spectal sequence to compute this and there is a multiplicative structure on that sequence I think I should be able to see how the cup product structure looks on the cohomology ring but I can't figure it out. I think it may have to do with the fact that the $E_2$ is isomorphic to $H^*(F)\otimes H^*(S^k)$.

Answer 1

I will quickly answer this myself as it is just a stupid mistake (would it be better to delete the question?). The actual homology of the fiber $F$ is \begin{align*} H^i(F)=\begin{cases} \mathbb{Z} & \text{if } i=0\\ \mathbb{Z}/n & \text{if } i=k+j(k-1)\\ 0 & \text{else} \end{cases} \end{align*} where $j\in\mathbb{N}_0$. Then the cupproduct structure will collapse for degree reasons.