A question about covering space Let $p: T \to X$ be a covering and let $f:Y\to X $ be a continuous function
we define $f^*T$ as
$$
f^*T=\{(y,\tilde{x})\in Y\times T|f(y)=p(\tilde{x})\}
$$
let $p':f^*T\to Y$ be the map given by $p'(y,\tilde{x})=y$
a) Prove that $p':f^*T\to Y$ is a covering.
b) Assume $(y_0,\tilde{x_0})\in f^*T $, prove that $p'_*(\pi_1(f^*T,(y_0,\tilde{x_0})))=f_*^{-1}(p_*(\pi_1(T,\tilde{x_0})))$ where $f_*:\pi_1(Y,x_0)\to\pi(X,f(x_0))$ is the homomorphism induced by $f$.
I think for a) 
assume $A$ is an open set and $A\subset Y$, then $p'^{-1}(A)=\{(y,\tilde{x})\in Y\times T|f(y)=p(\tilde{x}), y\in A\}$, next I hope to divide $p'^{-1}(A)$  into disjoint open union, and then prove every slice is homeomorphism to $A$, but I do not know how to divide $A$.
I think for b)
We have a theorem 

An element $[\omega]\in \pi_1(X,x_0)$ is in the subgroup $p_*(\pi_1(T,\tilde{x_0}))\subset \pi(X,x_0)$ if and only if the lift $\tilde{\omega}$ such $\tilde{\omega}(0)=\tilde{x_0}$ is a loop in $T$.

so $[\omega]\in p'_*(\pi_1(f^*T,(y_0,\tilde{x_0})))$ means for every of $\omega$ such that $\tilde{\omega}(0)=(y_0,\tilde{x_0})$ is a loop.
Also, $[\omega]\in f_*^{-1}(p_*(\pi_1(T,\tilde{x_0})))$ means for every lift of $\omega$
such that $\widetilde{f\omega}(0)=\tilde{x_0}$ is a loop.
But I do not know how to prove the two statements above are equivalent.
Thank you for your solution.
 A: For (a) you want to prove that every $y\in Y$ has an evenly covered neighborhood in $Y$.
Let's start with what we are given. We are given a continuous map $f:Y\to X$. So let $x=f(y)$. Now we are also given a covering map $p:T\to X$. So this means $x$ has an evenly covered neighborhood, call it $U$ and let $p^{-1}(U)=\bigsqcup_\alpha U_\alpha$. Let $V=f^{-1}(U)$. We would like $V$ to be an evenly covered neighborhood of $y$. We can see that $$p'^{-1}(V)=V\times_U\bigsqcup_\alpha U_\alpha=\bigsqcup_\alpha(V\times_U U_\alpha)$$ where each $V\times_U U_\alpha:=\{(z,\tilde x)\in V\times U_\alpha\ |\ f(z)=p(\tilde x)\}$ is open in $f^*T$.
I will leave it for you to check that the restriction to each set a  homeomorphism onto $V$. Bijectivity and continuity should be straight forward. You can show that $p'$ is an open map using the fact that $p$ is an open map.
Part b -
For part (b) first let $f':f^*T\to T$ be the map $(y,\tilde x)\mapsto \tilde x$ and observe that $f\circ p'=p\circ f'$ which gives that $f_*\circ p'_*=p_*\circ f'_*$
For simplicity, let us denote $G=\pi_1(f^*T,(y_0,\tilde{x_0}))$ and $H=\pi_1(T,\tilde{x_0})$. So we need to show that $p'_*(G)=f^{-1}_*(p_*(H))$
Let $p'_*([w])=[p'\circ w]\in p'_*(G)$. Then, $$f_*[p'\circ w]=[f\circ p'\circ w]=[p\circ f'\circ w]=p_*[f'\circ w]\in p_*(H)$$
For the converse, let $[v]\in \pi_1(Y,y_0)$. Then since $p'$ is a covering map, there is a unique lift of $v$ say $u:I\to f^*T$ beginning at $(y_0,\tilde x_0)$. Then $[v]=[p'\circ u]$ so that $[v]\in p'_*(G)$. Since $f^{-1}_*(p_*(H))\subseteq \pi_1(Y,y_0)\subseteq p'_*(G)$ we are done.
