Why is $\operatorname{Hom}_X(\widetilde{X},Y)\simeq p_Y^{-1}(x)$ for $Y,\widetilde{X}$ covering spaces of $X$ and $\widetilde{X}$ universal?

Question

Assume $X$ is a topological space that admits a universal cover $p: \widetilde{X}\rightarrow X$. Let $p_Y:Y\rightarrow X$ be another covering space of $X$. Since $\widetilde{X}$ is universal, there exists $f: \widetilde{X}\rightarrow Y$ such that $p_Y\circ f=p$. Let $\operatorname{Hom}_X(\widetilde{X},Y)$ denote the morphisms from $\widetilde{X}\rightarrow Y$ that commute with projections to $X$.

I have seen in various sources the claim that

$$\operatorname{Hom}_X(\widetilde{X},Y)\simeq p_Y^{-1}(x)$$

Why is this the case? This amounts to choosing a point $x_Y$ in the fiber $p_Y^{-1}(x)$. I see how $f(\widetilde{x})=x_Y$ for a $\widetilde{x}\in p^{-1}(x)$ and a choice of $x_Y\in p_Y^{-1}(x)$, but I don't see why this fixes the rest of the morphism $f$.

The rest of the morphism is determined by projecting down to $X$ and then lifting back up to $Y$, I think. — John Palmieri, Jun 12 '23 at 23:55
@JohnPalmieri not sure I follow, don't we still need to choose the lift in Y we lift up to — Tejas Rao, Jun 13 '23 at 02:36
See the lifting lemma as stated, for example, in https://math.stackexchange.com/a/145068/42781: the lifting is unique. — John Palmieri, Jun 13 '23 at 05:44

score 1 · Accepted Answer · answered Jul 26 '23 at 09:16

Universal coverings are usually only considered for connected locally path connected spaces $X$ and are understood as coverings $p : \tilde X \to X$ with a simply connected $\tilde X$. There is also a more general interpretation (see my answer to The universal cover covers any connected cover), but we shall not use it here.

Let us first consider any map $p : \tilde X \to X$ living on a connected space $\tilde X$ and define $$\operatorname{Hom}_X(\widetilde{X},Y) = \{ f : \tilde X \to Y \mid p_Y \circ f = p \} .$$ The elements of $\operatorname{Hom}_X(\widetilde{X},Y)$ are nothing else than the lifts of $p : \tilde X \to X$.

Fix a basepoint $x \in p(\tilde X) \subset X$ and choose $\tilde x \in p^{-1}(x)$. Since $\tilde X$ is connected, two lifts $f, f'$ of $p$ agree if and only if they agree at the point $\tilde x$ (i.e $f(\tilde x) = f'(\tilde x)$). For each lift $f$ of $p$ we have $f(\tilde x) \in p_Y^{-1}(x)$. Thus we get an injection $$\lambda : \operatorname{Hom}_X(\widetilde{X},Y) \to p_Y^{-1}(x), \lambda(f) = f(\tilde x) .$$

Now assume that $\tilde X$ is connected and locally path-connected. The well-known lifting theorem says the following:

Let $y \in p_Y^{-1}(x)$. Then a lift $f : \tilde X \to Y$ with $f(\tilde x) = y$ exists if and only if $p_*(\pi_1(\tilde X,\tilde x)) \subset (p_Y)_*(\pi_1(Y,y))$.

In other words, $y \in \lambda(\operatorname{Hom}_X(\widetilde{X},Y) )$ if and only if $p_*(\tilde X,\tilde x)) \subset (p_Y)_*(\pi_1(Y,y))$.

Therefore $\lambda$ is always a bijection if $\tilde X$ is simply connected and locally path-connected. Note that we do not need that $p$ is a covering map. However, if $p$ is a covering map, then we can conclude that

$\tilde X$ is locally path connected. In other words, if $p : \tilde X \to X$ is a covering map with a simply connected $\tilde X$ (which means that $p$ is a universal covering), then $\lambda$ is a bijection.
$p$ is surjective, thus we can take any $x \in X$ as a basepoint.

Why is $\operatorname{Hom}_X(\widetilde{X},Y)\simeq p_Y^{-1}(x)$ for $Y,\widetilde{X}$ covering spaces of $X$ and $\widetilde{X}$ universal?

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