Proving space is semi-locally simply connected Let $X$ be connected and locally path connected. Suppose $X$ has a covering map $p:\tilde{X} \rightarrow X$ such that $\pi_1(\tilde{X};\tilde{x_0})$ is the trivial group for some basepoint $\tilde{x_0} \in \tilde{X}$.
 I would like to show that this implies that $X$ is semi-locally simply connected.
 So I need to show that for every $x \in X$ and any neighbourhood $V$ of $x$ there exists an open subset $U \subset X$ such that $x \in U$ and the homomorphism $i_*:\pi_1(U;x) \rightarrow \pi_1(X;x)$ induced by the inclusion map $i:U \rightarrow X$ is trivial.
 I am not really sure where to start. Do I need to use the lifting criterion?
 A: Assuming $\tilde X$ is connected. So your condition shows that $ \pi_1(\tilde X,\tilde x)=0 \ \forall \ \tilde x\in \tilde X$.
I shall show if $x_0\in X$ and $\tilde x_0$ lies in the fiber over $x_0$ of $\tilde X\xrightarrow{p}X$, then $ \pi_1(\tilde X,\tilde x_0)=0\implies X$ is semi-locally simply connected at $x_0$.
Choose an open path-connected  neighbourhood $U\ni x_0$ and let $\tilde U\ni \tilde x_0$ such that $p:\tilde U\rightarrow U$ is an isomorphism. Then we have the commutative diagram
$$\require{AMScd}
\begin{CD}
\tilde U @>\tilde i>> \tilde X\\
@Vp|_{\tilde U}VV @VpVV \\
U @>i>> X
\end{CD}      $$
Applying $\pi_1$ we get the following commutative diagram
$$\require{AMScd}
\begin{CD}
\pi_1(\tilde U,\tilde x_0) @>\tilde i_*>> \pi_1(\tilde X,\tilde x_0)\\
@V(p|_{\tilde U})_*VV @Vp_*VV \\
\pi_1(U,x_0) @>i_*>> \pi_1(X,x_0)
\end{CD}      $$
Thus we get from the commutativity $i_*(p|_{\tilde U})_*=p_*\tilde i_*=0$ since $\pi_1(\tilde X,\tilde x_0)=0$
Since $(p|_{\tilde U})_*$ is an isomorphism, we get $i_*=0$
Applying this argument to other points completes the proof.
Edit: For the general case, let $\tilde X=\bigsqcup_i \tilde X_i$  be the connected components of $\tilde X$. Say $\tilde x_0\in \tilde X_{i_0}$ Then $p|_{\tilde X_{i_0}}:\tilde X_{i_0}\rightarrow X$ is a covering map with $\tilde X_{i_0} $ connected and you are back to the previous case.
