Prove that if $ p : Y → X$ is a locally trivial fibration with contractible fiber $Z$, and $X$ is a simplicial complex, then $Y$ is homotopy equivalent to $X$.
The previous exercise was to show that if $Y$ is contractible, then $X × Y$ is homotopy equivalent to $X$, for any topological space $X$. In this exercise i don't know how to show that there is homotopy $H(y,t): I×Y → Y$ such that $H(y,0) = p^{-1} \circ p$ and $H(y,1) = id|_Y$.
Possibly useful hint #1: Assume $X,Y$ are connected, since local triviality means you can do this argument component-by-component. The long exact fiber sequence has $$\pi_n(Z)\to\pi_n(Y)\to\pi_n(X)\to\pi_{n-1}(Z)$$ so since $\pi_n(Z)$ is trivial for $n$, $\pi_n(Y)\cong\pi_n(X)$ for all $n$. Thus, $X$ and $Y$ are weakly homotopy equivalent. We know $X$ is simplicial (so CW). If we knew $Y$ were simplicial, then by the Whitehead theorem $X$ and $Y$ would be homotopy equivalent.
Possibly useful hint #2: take the local trivializations and use them to use some subdivision of $X$ so that you can think of $Y$ as having a decomposition into spaces $\Delta^n\times Z$, where $\Delta^n$ ranges over the simplicies of $X$. Then since $Z$ is contractible, you can deformation retract all these spaces simultaneously. The tricky part is to make sure the boundaries for the decomposition agree.