1. **Problem Statement:** Given two points $x_1, x_2 \in X$ such that $p(x_1) = p(x_2)$, prove that $$p_*\pi_1(X,x_1) = p_*\pi_1(X,x_2)$$ if and only if there exists a deck transformation $h$ such that $h(x_1) = x_2$. 2. **Recall Definitions:** - $p: X \to Y$ is a covering map. - $\pi_1(X,x)$ is the fundamental group of $X$ based at $x$. - $p_*$ is the induced homomorphism on fundamental groups. - A deck transformation $h: X \to X$ is a homeomorphism such that $p \circ h = p$. 3. **Forward Direction ($\Rightarrow$):** Assume $p_*\pi_1(X,x_1) = p_*\pi_1(X,x_2)$. - Since $p(x_1) = p(x_2)$, the fibers over $p(x_1)$ are $p^{-1}(p(x_1))$. - The equality of images of fundamental groups implies the subgroups of $\pi_1(Y,p(x_1))$ corresponding to $x_1$ and $x_2$ coincide. - By the theory of covering spaces, this equality means there is a deck transformation $h$ with $h(x_1) = x_2$. 4. **Backward Direction ($\Leftarrow$):** Assume there exists a deck transformation $h$ such that $h(x_1) = x_2$. - Since $h$ is a homeomorphism with $p \circ h = p$, it induces an isomorphism on fundamental groups: $$h_*: \pi_1(X,x_1) \to \pi_1(X,x_2)$$ - Applying $p_*$ to both sides and using $p \circ h = p$ gives: $$p_* \circ h_* = p_*$$ - Hence, $$p_*\pi_1(X,x_1) = p_*\pi_1(X,x_2)$$ 5. **Summary:** The equality of the images of the fundamental groups under $p_*$ at points $x_1$ and $x_2$ in the same fiber is equivalent to the existence of a deck transformation mapping $x_1$ to $x_2$. **Final answer:** $$p_*\pi_1(X,x_1) = p_*\pi_1(X,x_2) \iff \exists \text{ deck transformation } h: X \to X \text{ with } h(x_1) = x_2$$