Induced map between fundamental groups from covering map is injective Question:
Let $f : X \to Y$ be a continuous map and let $x \in X$, $y \in Y$ be such that $f(x) = y$. Then there is an induced map $f_* : \pi_1(X, x) \to \pi_1(Y, y)$ such that $f_*([\gamma]) = [f \circ \gamma]$.
If $p$ is a covering map, show that $p_*$ is injective.
My Ideas: I need to show for $\gamma, \gamma' \in \pi_1(X, x)$, that if $p \circ \gamma$ and $p \circ \gamma'$ are homotopic, then $\gamma$ and $\gamma'$ are homotopic.
I feel like I am suppose to use the path lifting lemma somehow, but I don't see exactly how to do this.
$p \circ \gamma$ is a path in $Y$, and so for any point $(p \circ \gamma)(t)$ there is an open neighborhood $U$ such that $p^{-1}(U)$ is disjoint union of open sets mapped homeomorphicly by $p$ onto $U$. $\gamma(t) \in p^{-1}(U)$. I was thinking I could someone use this to construct the homeomorphism between $\gamma$ and $\gamma'$ but again I am drawing a blank.
Can anyone help point me in a direction? Thanks!
 A: What you're looking for is the homotopy lifting property:

Let $p:X\to Y$ be a covering map, and let $f_t:Z\to Y$ be a homotopy, with $\tilde{f}_0:Z\to X$ a lift of $f_0$.  Then there is a unique homotopy $\tilde{f}_t:Z\to X$ of $\tilde{f}_0$ lifting $f_t$

Now, let $Z=I$, so that $\tilde{f}_0$ is a path in $X$.  Suppose that $p\circ \tilde{f}_0=f_0$ is trivial in $\pi(Y,p(x_0))$, so that we have a homotopy $f_t:I\to Y$ taking $f_0$ to the constant path $f_1$.  By the homotopy lifting property, this gives us a homotopy $\tilde{f}_t$ which takes $\tilde{f}_0$ to a lift of the constant path.  By uniqueness, a lift of the constant path in $Y$ is the constant path in $X$, so that in fact $\tilde{f}_0$ is trivial in $\pi(X,x_0)$.
A: Instead of working with two loops, assume $\gamma\in \pi_1(X,x)$ such that $p\gamma \simeq k_y$ (constant loop at $y$) by a homotopy $\gamma_t$ keeping the endpoints fixed. By the homotopy lifting property, there is a homotopy $\gamma'_t:I \to X$ starting at $\gamma$ such that $p\gamma'_t = \gamma_t$. 
Now prove that $\gamma'_t(0) = \gamma'_t(1) =x$ for all $t$, and that $\gamma'_1=k_x$ the constant loop at $x$. This show that $\ker p_* = 0$
