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Let $X$ be a topological space, with $\gamma : [0,1] \to X $ some path in $X$. Suppose $p : \hat{X} \to X$ is a covering map. Then, by the path-lifting lemma, we know that for each $\hat{x} \in p^{-1} ( \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to \hat{X}$ such that $\hat{\gamma}(0) = \hat{x}$.

Similarly, the homotopy lifting lemma asserts the existence of unique lifts for maps $I \times I \to X$.

It has previously been remarked that homotopies may be considered as paths in $C(X,Y)$ endowed with an appropriate topology (e.g. compact-open). Therefore, my question is:

Question: can the homotopy lifting lemma be deduced from the path-lifting lemma applied twice?

I offer an outline of a plausible proof, which I would struggle to verify as I am not well-acquainted with the compact-open topology (or any topology on $C(X,Y)$).

Let $H : I \times I \to X$ be our candidate map for lifting.

  1. Define $\gamma_i : [0,1] \to X, x \mapsto H(x,i)$ (for $i = 0,1$).
  2. These give unique lifts $\hat{\gamma_i} : [0,1] \to \hat{X}$.
  3. Consider $\gamma_i$ as elements of $C([0,1],X)$. Note that there is a path $\eta$ between them in this space, described by $H$.
  4. Prove that $C([0,1] ,\hat{X})$ is a covering space of $C([0,1],X)$.
  5. By the path-lifting lemma, there exists a unique lift of $\eta : [0,1] \to C([0,1],X)$ to $\hat{\eta}:[0,1] \to C([0,1],\hat{X})$.
  6. This implies a unique lift $\hat{H}:I \times I \to \hat{X}$ of $H$, concluding the proof.

The steps (1),(2) and (5) are consequences of the path-lifting lemma. (3) may be assumed given a suitable topology on $C(X,Y)$. Thus, this question boils down to proving that [there exists a topology on $C(X,Y)$ such that] steps (4) and (6) are valid.

Edit: a natural suggestion for the covering map in (4) is:

$$P : \hat{\gamma} \mapsto p \circ \hat{\gamma}$$

I am likely to have made a typographical error somewhere. Please do mention any you find.

legionwhale
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  • One of the problems is that passing from $I\to C(I,X)$ to $I\times I\to X$ and back requires some additional assumptions on $X$, typically locally compact Hausdorff. I don't think point (6) can be easily proven without it. – freakish Apr 11 '23 at 07:31
  • The homotopy lifting lemma asserts that if you have a homotopy $I\times I\rightarrow X$ and a partial lift $I\times{0}\rightarrow\hat{X}$, then there is a unique homotopy $I\times I\rightarrow\hat{X}$ that extends this partial lift and lifts the original homotopy. If you update your strategy to reflect this, it can be made to work. You need to know the currying properties of the compact-open topology (I disagree with @freakish, the only space that need be LCH here is $I$). The hard part is proving (4), which is slightly tricky. – Thorgott Apr 11 '23 at 10:46
  • @Thorgott The statement of the homotopy lifting lemma I know is the slightly more general one I gave. I appreciate that a map from $I \times I \to X$ is not necessarily a homotopy, but I assumed that it acquired that name from its most frequent use-case. Also, doesn't the existence of the lift $I \times {0} \to \hat{X}$ follow from the path-lifting lemma, as in (1)? Do you have any hints or would you mind writing an answer on how to make this work? I am not even familiar with currying in algebraic topology. – legionwhale Apr 11 '23 at 14:41
  • Rather than more general, your statement, if I'm extrapolating correctly, is simply not quite true. A map $I\times I\rightarrow X$ does not have a unique lift through $\hat{X}\rightarrow X$. The lift is only unique if you require it to extend a partial lift $I\times{0}\rightarrow\hat{X}$. This is analogous to path lifting. A path $I\rightarrow X$ does not have a unique lift through $\hat{X}\rightarrow X$. The lift is only unique if you require it to start at a fixed point in $p^{-1}(\gamma(0))$ (borrowing your notation). There can be multiple lifts with different starting points. – Thorgott Apr 11 '23 at 17:25
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    @Thorgott My bad, I could have assisted in your extrapolation; my statement assumes that the lift is required to "start" at some point in $p^{-1}(H(0,0))$ -- which is to say, just a point, as for the path-lifting lemma. Indeed, I don't see why we can't deduce the unique partial lift $I \times {0} \to \hat{X}$ exists starting from this point. – legionwhale Apr 11 '23 at 20:23
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    You are correct at that point, specifying the starting point or a partial lift on $I\times{0}$ is equivalent, but you do need some initial specification to begin with in order to get a unique lift at the end. I just find this version more convenient cause the partial lift corresponds to a point in $C([0,1],\hat{X})$ that functions as starting point when you try to apply unique path lifting to the covering map $C([0,1],\hat{X})\rightarrow C([0,1],X)$. Re: currying, try p.529 onwards in Hatcher's Algebraic Topology. – Thorgott Apr 12 '23 at 00:53
  • @Thorgott Thanks, I'll take a look. – legionwhale Apr 12 '23 at 02:25

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