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I read on the English Wikipedia page on covering spaces that “a covering space is a universal covering space if it is simply connected”, which looks like an actual definition of a covering space being a universal covering space.

However, I read in other sources other definitions of a covering space. For example the French Wikipedia page on covering spaces (Revêtement) uses definition 4), infra.

So, which one of these definitions is the “correct” one ?

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(1) The mapping $q : D \to X$ is a universal cover of the space $X$ if $D$ is simply connected ;

(2) The mapping $q : D \to X$ is a universal cover of the space $X$ if for any cover $p : C \to X$ of the space $X$ where the covering space $C$ is connected, there exists a covering map $f : D \to C$ such that $p \circ f = q$ ;

(3) The mapping $q : D \to X$ is a universal cover of the space $X$ if it is a Galois cover and for any cover $p : C \to X$ of the space $X$ where the covering space $C$ is connected, there exists a covering map $f : D \to C$ such that $p \circ f = q$ ;

(4) The mapping $q : D \to X$ is a universal cover of the space $X$ if it is a Galois cover and for any cover $p : C \to X$ of the space $X$, there exists a covering map $f : D \to C$ such that $p \circ f = q$.

It is true that $(1) \implies (3) \implies( 2)$ and $(4) \implies (3), (2)$.

Is it also true that $(1) \implies (4)$ ? $(2) \implies (1)$ ? $(3) \implies (1)$ ?

For example, using definitions 1), 2) or (3), it is clear that “the” universel cover of a point is itself, and the cover mapping $q : \{ \bullet \} \to \{ \bullet \}$ is the identity. Indeed :

Using definition 1), the point is simply connected, so its universal cover is itself.

Using definition 2) or 3), note that the identity map $q : \{ \bullet \} \to \{ \bullet \}$ is a Galois cover and that the covers of $\{ \bullet \}$ are all the non-empty discrete spaces, so the only connected ones are the 1-point spaces $C$ (the cover mapping being the natural bijection $p : C \to \{ \bullet \}$). Therefore there exists a covering map $f : \{ \bullet \} \to C$ (which is just $p^{-1}$) which satisfies $p \circ f = q$.

However, we can’t prove using definition 4) that “the” universel cover of a point is itself, since if $p : F \to \{ \bullet \}$ is a non-connected cover where $F$ is a discrete space of cardinality at least 2, there is no covering map $f : \{ \bullet \} \to F$ such that $p \circ f = q$ (as a covering map should be surjective).

So my final question is :

Is the universal cover of a 1-point space really a 1-point space ?

Thanks.

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It is a little hard to say “what definition is correct”, since there are some definitions which are more appropriate to be context-dependent (see (*)), but let me make a point.

A universal cover is essentially a solution to a universal property problem. That being its name (universal), it seems to me that the natural definition is the one that makes it really be… universal.

What universal property does it satisfy?

Consider the category $\mathcal{C}$ which has as objects covering maps from pointed topological spaces to a fixed pointed topological space $(X,x_0)$. That is, the objects are $p:(Y,y_0) \to (X,x_0)$ (more on this later). And the morphisms are covering maps $c: (Y_1,y_1) \to (Y_2,y_2)$ making the following diagram commute

$$\begin{array}{ccccccccc}

& & (Y_1,y_1) \\

& \swarrow{c} & \downarrow{p_2} \\

(Y_2, y_2)& \xrightarrow{p_1} & (X,x_0).

\end{array}$$

We want a *universal cover* to be a initial object in this category. Why? A universal cover should cover any cover, and that is precisely what the above property is capturing.

If we take the above assumption as a definition, how do we reconcile it with the usual definition as a simply connected cover? Well, consider the following theorem:

Theorem:[Lifting Theorem] Let $p:(Y,y_0) \to (X,x_0)$ be a covering map.Assume that$W$ is path-connected and locally path-connected and that $f: (W,w_0) \to (X,x_0)$ is a given map. Then, there exists a lifting of $f$ to $(Y,y_0)$ covering $f$ if and only if $f_{\#} \pi_1(W,w_0) \subset p_{\#}\pi_1(Y,y_0)$. Moreover, such a lifting is unique.

A moment of thought then tells us that if $\pi_1(W,w_0)$ is trivial in this theorem, then there always exists a lifting. Another moment of thought then shows us that if there is a object in our category $\mathcal{C}$ which has as the covering space a simply connected space, then it is the universal cover (with a small detail: we would need to prove that the lifting is indeed a covering map, but this is true).

Now, some remarks.

To guarantee uniqueness of things, we need to suppose that the spaces are pointed. Due to the hypothesis of the lifting theorem, we also need path-connectedness and locally path-connectedness. If we want a “clean” theory (that is, trying to minimize recurrent restatement of hypotheses), then a good solution is to clump up all of those things in a definition. For instance, that is what Bredon does in his topology and geometry book:

Definition:A map $p: X \to Y$ is called acovering map(and $X$ is called acovering spaceof $Y$) if $X$ and $Y$ are Hausdorff,arcwise connected, and locally arcwise connectedetc etc.

OBS: Note that “arcwise connected” means “path-connected” for Bredon.

Note now the phenomenon that I alluded to at the beginning: Bredon himself “changes” definition later for a specific purpose: On page $342$, he states:

“For convenience, we use the word ‘cover’ in Proposition $7.6$ to mean everything in the definition of a covering space except for the connectivity requirements.”

(*)So, this answers ends up not answering our question. But that is because I think that it is unanswerable: some definitions are better than others when on different context, needing less or more from them. The situation is similar to how some texts define compact as “compact and Hausdorff”, or some texts define “ring” with unity or not, or “regular space” entailing Hausdorff or not etc.

However, to answer your final question: Yes, the universal cover of a point is a point.

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