Reposted from this Reddit post as I didn't get good answers there:
So I've been reading about the Galois theory of covering maps and been staring at this equation for way too long:
$$\left| \pi_1(X,x_0)/p_*(\pi_1(\tilde{X},\tilde{x}_0))\right| = |p^{-1}(x_0)|$$
Where $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ is a covering map with $\tilde{X}$ path connected. And the subgroup is normal iff the covering map is Galois, and the quotient is isomorphic to the set of deck transformations, which is exactly singularly transitive on the fibers, etc.
So $p$, the covering map is famously surjective, famously not at all injective. And the induced homomorphism $p_*$ is the opposite: Injective, (by homotopy lifting) but not at all surjective. And this equation says that the degree that $p_*$ fails to be surjective (the cokernel) is exactly the same as the degree $p$ fails to be injective, at least at $x_0$. Is there a dual to this notion? Some sort of type of injective continuous map (so an embedding $\iota$), or topological condition on a subset that always induces a surjection $\iota_*$, so that $\text{ker}(\iota_*)$ is somehow related to how $\iota$ fails to be surjective? Am I getting anywhere with this?
Some rudimentary work:
Firstly I set myself four test examples: Two simple ones:
$$\iota:S^1\to S^2$$ $$\iota:S^1 \to S^1\times \mathbb{R}$$ In both we have that $\iota$ is injective with $\iota_*$ surjective, but in the former $\text{ker}(\iota_*) = \pi_1(S^1)$, and in the latter $\text{ker}(\iota_*) = \{e\}$. And a more complicated one: Let $X$ be any connected topological space with a finitely generated fundamental group and let $A$ be the graph of $\pi_1(X,x_0)$'s generators. $\iota$ is the inclusion map as usual. Then $\pi_1(A,x_0)$ would be the free group on $n$ elements, and the kernel would be generated by whatever conditions of the group. (My first formulation was $X$ a torus, with one curve inside the rim, and one around it making $A$ a lemniscate with one ended twisted $90$ degrees) And four, a non-example: the inclusion map of $S^1$ into the lemniscate, where $\iota_*$ is not surjective.
A sufficient condition for $\iota_*$ to be surjective would be a deformation retraction on to the subset. But it's not necessary, $\iota:S^1\to S^2$ doesn't have one but still has a surjective homomorphism. So I tried editing it to make it local: Something like $\iota:A\to X$ is a local deformation retract if for every $x\in X$, $x\in \mathcal{U}$ for some open nbhd $\mathcal{U}$ such that $\mathcal{U}$ has a deformation retract on to $\mathcal{U}\cap A$. But then this condition would cover example four, $S^1$ into the lemniscate where $\iota_*$ is not surjective.
Another attempt: A sort of idiot's version of a covering map is the projection $p:X\times D\to X$ where $D$ is a discrete space, where the cardinality of $D$ will be the cardinality of the equation on top. All covering maps look like this locally. A friendly redditor said that the right dualization would be $\iota: X\to X\sqcup D$, but that would make $\iota_*$ an isomorphism always (even if you "localize" it, I think). I think a closer attempt would be to go from $p:\sqcup_{\alpha} X\to X$ to $\iota: X\to \prod_{\alpha} X$, or again a local version: $\iota: X\to X'$ such that for every $x'\in X'$, there exists a nbhd $\mathcal{U}$ of $x'$ such that $\mathcal{U}$ is homeomorphic to $\prod X\cap \mathcal{U}$. This would work for the first two examples where $\mathcal{U}$ could be a sort of open rectangle chart and $X\cap \mathcal{U}$ would be homeomorphic to an open interval, but not at all the third example. There if we take $x'$ to be the basepoint $\mathcal{U}$ would also be homeomorphic to a open set in $\mathbb{R}^2$ but $\mathcal{U}\cap X$ would be some sort of X shape. Another thought: Maybe we should also dualize $D$ from discrete space to a trivial topology.
Another redditor mentioned cofibrations and the homotopy extension property, but that seems to cover maps from the space, and not maps to it, like paths. My knowledge is limited when it comes to that.
