# Simply connected means universal-covering spaces

A covering space $p:(Y,y_0) \to (X,x_0)$ is called universal when $Y$ is simply connected (say that we restrict ourselves to path connected spaces, and locally path connected). I heard that this definition is natural since whether you have an another covering space $q:(Z,z_0) \to (X,x_0)$ then $(Y,y_0)$ is a covering of $(Z,z_0)$. I know that this should follow from the following lifting criterion

Suppose that we have covering space $q:(Z,z_0) \to (X,x_0)$ and a map $f:(Y,y_0) \to (X,x_0)$ where $Y$ is path connected and locally path connected. Then there is a lift $\tilde{f}:(Y,y_0) \to (Z,z_0)$ of $f$ iff $f_*(\pi_1(Y,y_0)) \subset p_*(\pi_1(Z,z_0))$.

If now $Y$ is simply connected then this condition is satisfied and therefore if we take $f$ to be the covering map $p$ then we can lift it to $\tilde{p}:(Y,y_0) \to (Z,z_0)$. My quess would be that $\tilde{p}:(Y,y_0) \to (Z,z_0)$ would be the desired covering but I don't see how to get this.

EDIT: according to the answer below: I found the result which you used in Spanier's book (Chapter 2 p. 64). First he shows (Lemma 12) that if we have an open and connected subset $U \subset X$ which is evenly covered by $p:\tilde{X} \to X$ (meaning that it is a disjoint union of open sets all of them homeomorhic with $U$ via $p$) then $p$ maps each component of $p^{-1}(U)$ homeomorphically onto $U$. This is fine and below we have a corollary of this fact which is precisely the result which you have used. The proof goes as follows: if $U$ is connected open subset of $X$ evenly covered by both $p_1$ and $p_2$ then it is easy to deduce from the above result that each component of $p_2^{-1}(U)$ is evenly covered by $p$. I tried to see how this argument goes and my attempt is the following: suppose that $p_1^{-1}(U)=\bigsqcup_iV_i$ and $p_2^{-1}(U)=\bigsqcup_jW_j$. From Lemma 12 we know that each $W_j$ is a component. Now I consider the restricted diagram between $p^{-1}(W_j)$, $W_j$ and $U$ (arrows are $p|_{p^{-1}(W_j)},p_2|_{W_j},p_1|_{p^{-1}(W_j)}$ and it follows that $p^{-1}(W_j)=\bigsqcup_i (V_i \cap p^{-1}(W_j))$. Is it enough? Is it true now that $p$ maps homeomorphically each of $V_i \cap p^{-1}(W_j)$ onto $W_j$?

Suppose that we fix a connected locally path connected space $X$ and two covering projections $p_i: X_i\to X$, $i=1,2$ with $X_i$ connected, and $f:X_1\to X_2$ is a continuous map making the obvious diagram commute. Then $f$ is a covering projection if $f$ is surjective, so all you have to prove is that this holds assuming $X$ is connected locally connected. Recall that if $X$ is locally path connected so are the $X_i$, and since they are connected by hypothesis, they are path connected. Given $x_2\in X_2$, take $x_1\in X_1$ and a path $\omega_2$ from $f(x_1)$ to $x_2$. Consider now the path $p_2\omega_2$ in $X$. Since $p_1$ is a covering, there is a path $\omega_1$ starting at $x_1$ in $X_1$ such that $p_1\omega_1=p_2\omega_2$. But then $p_1=p_2f$ gives $f\omega_1=\omega_2$ by the unique lifting property. In particular $f\omega_1(1)=\omega_2(1)=x_2$, so $f$ is onto.