Unique lifting property

Statement

Let $(X,\mathcal{ J })$ be a topological space, suppose that $(E,\mathcal{ H })$ is a covering space (with covering map $p:E\rightarrow X$). Suppose $(Y,\mathcal{ K })$ is a connected topological space and $f:Y\rightarrow X$ is a continuous map, then^{Partial: -}

• given two lifts of $f$ through $p$, say $g,h:Y\rightarrow E$ we have:
  • $\big(\exists y\in Y[g(y)\eq h(y)\big)\implies \big(\ \underbrace{\forall y\in Y[g(y)\eq h(y)]}_{\text{i.e. that }g\eq h}\ \big)$
    • In words: if there exists a point on which $g$ and $h$ agree then $g$ and $h$ are equal as functions

Bonus corollary

Recall that a logical implication is logically equivalent to the contrapositive, that is

• $(A\implies B)\iff(\neg B\implies\neg A)$

So, should the above claim be true, we also get:

• $\big(\exists y\in Y[g(y)\neq h(y)]\big)\implies\big(\forall y\in Y[g(y)\neq h(y)]\big)$
  • In words: if there exists a $y_0\in Y$ such that $g$ and $h$ disagree at $y_0$ then they disagree everywhere.
    • Caveat:This does not mean $g(Y)\cap h(Y)\eq\emptyset$ necessarily!

Proof

Grade: C

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Warning:What follows is VERY messy. I was distracted when writing it and kept switching between it and something else. However the written proof was BARELY legible so this is better than nothing. Hopefully

Warning:That grade doesn't exist!

TODO: It'd be good to put some reasoning here. Like given a $z\in Y$ to show $g$ and $h$ agree there we really need to work in terms of open neighbourhoods to $z$, which we can get from somewhere they're known to agree on. However this only works if it's connected, as if $z$ is in another connected components "stuff just doesn't work" <--make this more formal

Let us make the following definitions:

• $S:\eq\{y\in Y\ \vert\ h(y)\eq g(y)\}$ (when we introduce the hypothesis, by that hypothesis this will be non-empty)
• $T:\eq\{y\in Y\ \vert\ h(y)\neq g(y)\}$

Lemma 1: $S$ and $T$ are disjoint.

• Suppose $S\cap T\neq\emptyset$, clearly this $iff\exists z\in S\cap T$
  • Suppose there is such a $z\in S\cap T$, by definition of intersection that means $\exists z\in Y[z\in S\wedge z\in T]$
    • $z\in S\iff g(z)\eq h(z)$ and $z\in T\iff g(z)\neq h(z)$
      • We see $g(z)\eq h(z)\neq g(z)$ so $g(z)\neq g(z)$ - a contradiction!
• We see there cannot be any elements in $S\cap T$ for if there was we have a contradiction. As shown
• So $S\cap T\eq\emptyset$ - i.e. $S$ and $T$ are disjoint

Lemma 2: $S\cup T\eq Y$ AND $S\eq Y-T$ (and $T\eq Y-S$) in some form. We need $S\eq Y-T$ and $T\eq Y-S$ at least!

TODO: Todo

As $Y$ is a connected topological space we see that the only sets that are both open and closed are $Y$ itself and $\emptyset$, if the result holds (which we very much hope it does) then $S\eq Y$ and $T\eq\emptyset$, so if we show $S$ is open. As $T\eq Y-S$ we would see $T$ is closed as a result.

If we show $T$ is open too, then $S$ would be closed, thus showing they're both open and closed!

So we must have $S\eq Y$ (as $S$ is non-empty by hypothesis) and $T\eq\emptyset$, but if $S\eq Y$ then they agree everywhere!

TODO: Notes:Covering spaces shows it better

Proof that $S$ is open

• Let $s\in S$ be given. There is at least one to give by hypothesis. We must find a neighbourhood of $s$ that is fully contained in $S$, recall that a set is open if and only if it is a neighbourhood to all of its points, this is what we hope to show.
  • Define $r:\eq h(s)\eq g(s)$ and notice $r\in E$
    • Define $z:\eq p(r)$ and notice $p(r)\in X$
      • As $z\in X$ and $p$ is a covering map there exists an open neighbourhood, $U\in\mathcal{J}$ of $z$ such that $U$ is evenly covered by $p$
      • Choose $U$ to be such an open neighbourhood
        • By choice of $U$ we see $\exists(V_\alpha)_{\alpha\in I}\subseteq\mathcal{H}$
          • such that:
            1. $p^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha$
            2. The $(V_\alpha)_\alpha$ are open (given by being in $\mathcal{H}$)
            3. The $(V_\alpha)_\alpha$ are pairwise disjoint
            4. For each $V_\beta\in(V_\alpha)_\alpha$ we have $V_\beta\cong_{p\vert^\text{Im}_{V_\beta} } U$, that is the covering map restricted to its image on $V_\beta$ is a homeomorphism onto $U$
        • Choose $(V_\alpha)_{\alpha\in I}\subseteq\mathcal{H}$ to be this family of sheets of the covering
          • As $z\in U$ we see $p^{-1}(z)\subseteq p^{-1}(U)$
          • As $z:\eq p(r)$ we see that $r\in p^{-1}(z)$ so $r\in p^{-1}(U)$
          • As $p^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha$ we see $r\in \bigcup_{\alpha\in I}V_\alpha$, by definition of union:
            • $\big(r\in \bigcup_{\alpha\in I}V_\alpha\big)\iff\big(\exists\beta\in I[r\in V_\beta\big)$
          • So $\exists\beta\in I[r\in V_\beta]$
            • Define $V:\eq V_\beta$ where $V_\beta$ is the element of $(V_\alpha)_{\alpha\in I}$ with $r\in V_\beta$ as discussed above
              • As $r:\eq h(s)$ and $h(s)\eq g(s)$ we see that:
                • $r\in V\iff[h(s)\in V\wedge g(s)\in V]$
              • So $s\in h^{-1}(V)$ and $s\in g^{-1}(V)$
                • Notice, by continuity and $V$ being open in $(E,\mathcal{ H })$ that $h^{-1}(V)$ and $g^{-1}(V)$ are both open in $(Y,\mathcal{ K })$.
              • Thus $s\in h^{-1}(V)\cap g^{-1}(V)$ (as by definition of intersection $(s\in A\cap B)\iff(s\in A\wedge s\in B)$ - we have the RHS, so we have the left.
                • Define $W:\eq h^{-1}(V)\cap g^{-1}(V)$, so $W\subseteq Y$
                  • notice:
                    1. that $W$ is open in $(Y,\mathcal{ K })$ as by definition of a topology the intersection of open sets is open.
                    2. $W\neq\emptyset$ ($W$ is non-empty) as - at least - $s\in W$
                  • Define $q:\eq p\vert_V^\text{Im}:V\rightarrow U$ be the homeomorphism of the restriction of $p$ to $V$ which is onto $U$.
                    • This means that $q$ is injective, i.e.:
                      • $\forall v,w\in V[q(v)\eq q(w)\implies v\eq w]$
                    • Notice also that $h(W)\subseteq V$ and $g(W)\subseteq V$
                    • Let $w\in W$ be given
                      • By definition of being lifts: $f(w)\eq p(h(w))\eq p(g(w))$
                      • As $h(w)\in V$ and $g(w)\in V$ we see that $p(h(w))\eq q(w)$ and $p(g(w))\eq q(g(w))$
                        • So $f(w)\eq q(h(w))\eq q(g(w))$
                          • But $q$ is injective, so $q(h(w))\eq q(g(w))\implies g(w)\eq h(w)$
                      • so we have $g(w)\eq h(w)$
                        • so $w\in S$ (by definition of $S$)
                    • Since $w\in W$ was arbitrary we have shown $\forall w\in W[w\in S]$
                • Thus $W\subseteq Y$, that is $Y$ contains a neighbourhood of $s$
• Since $s\in S$ was arbitrary, we have shown $\forall s\in S\exists W\in\mathcal{K}[x\in W\wedge W\subseteq Y]$
  • In words: that for all points in $S$ there is a neighbourhood to that point contained entirely in $S$

Thus $S$ is open, as required.

Grade: A

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We still need to show $T$ is open