Problem 4.1.6 in Hatcher's Algebraic Topology book says:
Suppose that $p : (\tilde{X}, \tilde{A}, \tilde{x_0}) \to (X,A,x_0)$ is a covering space with $\tilde{A} = p^{-1}(A)$. Show that the induced map $p_* : \pi_n((\tilde{X}, \tilde{A}, \tilde{x_0}) \to \pi_n(X,A,x_0)$ is an isomorphism for $n > 1$.
Using the long exact sequence of the pair, this is clear for $n > 2$. For $n = 2$, it is easy by doing some diagram chasing to show that $p_*$ is injective. However, it is not clear to me that it is surjective.
I believe I have a counter example:
Edit: This example doesn't work, as is pointed out in comments. I'm still not sure why this is true though.
Let $X = \bar{\mathbb{D}}^2 = \tilde{X}$, the closed unit disc, and $A = S^1 = \partial X$. Let $p : \tilde{X} \to X$ be the squaring map. Then using the naturality of the long exact sequence of the pair, the map $\pi_2(\tilde{X}, \tilde{A})$ agrees with $\pi_1(\tilde{A}) \to \pi_1(A)$, which is multiplication by $2$ from $\mathbb{Z} \to \mathbb{Z}$. This map is not surjective.
Maybe I'm missing something -- or perhaps this is a typo in the text? I don't see it in the errata list, and the statement of the problem is from most recent version.
For surjectivity at $p_*\colon \pi_2(\tilde{X},\tilde{A},\tilde{x_0}) \to \pi_2(X,A,x_0)$ note that an element of $\pi_2(X,A,x_0)$ can be represented by a map $$ x\colon (\Bbb D^2, \Bbb S^{1},s_0)\to (X,A,x_0) $$ since $\Bbb D^2$ is simply connected this map can be lifted to $\tilde{X}$, so we have a unique map $$\tilde{x} \colon (\Bbb D^2,s_0) \to (\tilde{X}, \tilde{x_0}) $$ with $p \circ \tilde{x} = x$. We have $\tilde{x} (\Bbb S^1) \subset \tilde A $ and thus $\tilde{x}$ represents a presage of $x$. Hence the maps is surjective.