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?

• Is $\tilde X$ is connected ? Commented Oct 23, 2020 at 16:09
• It's not mentioned in the problem. Commented Oct 23, 2020 at 16:12

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.