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I was thinking about how $S^1 \hookrightarrow T^2$ and how the universal cover of $S^1$ which is $\mathbb{R}$ is contained in the universal cover of $T^2$ which is $\mathbb{R}^2$. Is this a coincidence or does this always hold for topological spaces? That is, if $X$ embeds into $Y$, then $\widetilde{X}$ embeds into $\widetilde{Y}$? Or at least, there is some injective map? Let's assume that the spaces are path connected and locally path connected.

One thought I had seems to get me close to this result. Let $p:\widetilde{X} \to X$ be a covering map and $i:X \hookrightarrow Y$ an embedding.Then $f = i \circ p: \widetilde{X} \to Y$. This map, $f$ lifts to a map $\tilde{f}:\widetilde{X} \to \widetilde{Y}$ if and only if $f_*(\pi_1(\widetilde{X})) \subset p_*(\pi_1(\widetilde{Y}))$ (these are induced homomorphisms on fundamental groups). Well, $\pi_1(\widetilde{X}) = \pi_1(\widetilde{Y}) = 0$ by definition of universal covers so there exists such a $\tilde{f}$.

But I'm not sure if this map is a topological embedding; i.e. homeomorphism onto its image. Here are my thoughts. $f$ is not 1-1 but it is at least onto the image of $i$ which is basically a copy of $X$ sitting in $Y$. So restricting our maps to $f:\widetilde{X} \to i(X) \cong X$ and $\tilde{f}:\widetilde{X} \to \widetilde{Y}$, we know that since this altered $f$ is onto and $p$ is onto, we need $\tilde{f}$ to "hit enough" points in $p^{-1}(Y)$ so that they get projected onto $Y$. But $\widetilde{X}$ is connected and $\tilde{f}$ is continuous. Thus, $\tilde{f}(\widetilde{X}) \subset \widetilde{Y}$ should be connected. Covering maps have discrete fibers so $\tilde{f}(\widetilde{X})$ should sort of be a single slice or at least connected slices.

I'm not sure if any of that is useful information.

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    $\begingroup$ Note that the circle is contained contained in the Hawaiian Earing, but this space has no universal cover. $\endgroup$
    – Aweygan
    May 8 '18 at 2:56
  • $\begingroup$ Nevertheless, you can strengthen the hypotheses by assuming both $X$ and $Y$ have universal covers, in which case I have not thought of a solution yet. $\endgroup$
    – Aweygan
    May 8 '18 at 2:59
  • $\begingroup$ My suggestion would be to assume that the induced map is injective on $\pi_1$. $\endgroup$
    – Thomas Rot
    May 8 '18 at 12:44
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No. For instance, the circle, $S^1$, embeds as the boundary of the disk $D^2$, but the lift $\mathbb{R}^1 \to D^2$ is not an embedding.

That said, if you have additional structure, such as nonpositive curvature, then what you observe is correct. Suppose $X \to Y$ is a locally isometric embedding and $Y$ is nonpositively curved, e.g. a complete locally CAT(0) metric space. Then the lift $\widetilde{X} \to \widetilde{Y}$ is an isometric embedding. The idea is that if $c:[0,L] \to \widetilde{X}$ is an isometric embedding, joining $x_0 = c(0)$ to $x_1 = c(L)$ so that $d_X(x_0, x_1) = L$, then $c$ projects to a local geodesic in $X$ and over to a local geodesic in $Y$, which then lifts to a local geodesic in $\widetilde{Y}$. But since $\widetilde{Y}$ is globally nonpositively curved, e.g. CAT(0) (note I've dropped the "locally"), this local geodesic is a global geodesic, meaning that the images of $x_0$ and $x_1$ in $\widetilde{Y}$ are exactly distance $L$ apart.

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