If $H \le \pi_1(X,x)$ is conjugate to $P_*(\pi_1(Y, y))$, then $H \cong P_*(\pi_1(Y, y'))$ for some $y' \in P^{-1}(x)$ Assuming $P: Y \to X$ is a covering map, I want to show that if $H \le \pi_1(X,x)$ is conjugate to $P_*(\pi_1(Y, y))$, then $H \cong P_*(\pi_1(Y, y'))$ for some $y' \in P^{-1}(x)$.
I started off with letting $H = [g]^{-1} * P_*(\pi_1(Y, y)) * [g]$, where $g \in \pi_1(X, x)$ but am having trouble determining where each element of $H$ is mapped to in $P_*(\pi_1(Y, y'))$.
 A: Let $\gamma$ be a loop based at $x$ in $X$ such that $H=[\gamma]P_*(\pi_1(Y, y))[\gamma]^{-1}$. Let $\tilde \gamma:I\to Y$ be the lift of $\gamma$ starting at the point $y\in Y$. Let $y'=\tilde \gamma(1)$, that is, $y'$ be the end point of $\tilde\gamma$. We will show that $H=P_*(\pi_1(Y, y'))$.
To this end, let $h\in H$ be arbitrary. Then there is a loop $\tilde \theta$ based at $y$ in $Y$ such that $h=[\gamma][\theta][\gamma]^{-1}$, where $\theta=P\circ \tilde\theta$. Consider the loop $\tilde\alpha:=\tilde \gamma*\tilde \theta*\bar{\tilde \gamma}$ based at $y'$ in $Y$. Then $P\circ\tilde\alpha$ is $\gamma*\theta*\bar\gamma$, and thus $P_*([\tilde\alpha])= [\gamma][\theta][\gamma]^{-1}$, showing that $H$ is in the image of $P_*:\pi_1(Y, y')\to \pi_*(X, x)$. But every loop based at $y'$ in $Y$ can be wrttien as $\tilde\gamma*\tilde \beta*\bar{\tilde \gamma}$ for some loop $\tilde \beta$ based at $y$.
(This is because of $X$ is any topological space and $\alpha$ is a path in $X$ joining $p$ to $q$, then there is an isomorphism $\alpha_*:\pi_1(X, p)\to \pi_1(X, q)$ which is given by "conjugating with $\alpha$".)
Thus the image of $P_*:\pi_1(Y, y')\to \pi_1(X, x)$ is also contained in $H$, showing that $P_*(\pi_1(Y, y'))= H$. In fact $P_*$ maps $\pi_1(Y, y')$ isomorphically onto $H$ because $P_*$ is injective.
