Return to Answer

The relationship goes in 2 steps. First the fundamental pro-group can be compared with the Cech fundamental group endowed with the inverse limit topology. In particular, as noticed by Atiyah and G. Segal, they contain precisely the same information as long as the fundamental pro-group is Mittag-Leffler. For other results in this direction see lemma 3.4 in "Steenrod homotopy". The topological homomorphism between $\pi_1(X)$ (with topology as in my comment) and the topological Cech fundamental pro-group is discussed in theorem 6.1, section 5 and elsewhere in "Steenrod homotopy".

The relationship goes in 2 steps. First the fundamental pro-group can be compared with the Cech fundamental group endowed with the inverse limit topology. In particular, as noticed by Atiyah and G. Segal, they contain precisely the same information as long as the fundamental pro-group is Mittag-Leffler. For other results in this direction see lemma 3.4 in "Steenrod homotopy". The topological homomorphism between $\pi_1(X)$ (with topology as in my comment) and the topological Cech fundamental pro-group is discussed in theorem 6.1, section 5 and elsewhere in "Steenrod homotopy".

The Mathoverflow thread on the quotient topology on $\pi_1$, and particularly Andrew Stacey's answer there, make it clear that the only reason that this topology is not compatible with multiplication is that the product of two quotient maps need not be a quotient map. But wait, the product of two quotient maps is a quotient map in the category of uniform spaces and uniformly continuous maps! See Isbell's "Uniform spaces" (1964), Exercise III.8(c). In fact, the product of any (possibly infinite) collection of quotient maps is a quotient map in the uniform category. (Quotient maps are defined in any concrete category over the category of sets, as explained e.g. in The Joy of Cats.)

Added still later: With this in mind, I no longer see a good reason for myself to think at all about the quotient topology on $\pi_1(X)$. As you can see for instance from my comment here, I have good reasons to believe that quotient topology is a sensible notion only when it coincides with the topology of quotient uniformity (i.e. for quotients of compact spaces), and otherwise the topology of quotient uniformity is the way to go. The present situation does not seem to be an exception.

The Mathoverflow thread on the quotient topology on $\pi_1$, and particularly Andrew Stacey's answer there, make it clear that the only reason that this topology is not compatible with multiplication is that the product of two quotient maps need not be a quotient map. But wait, the product of two quotient maps is a quotient map in the category of uniform spaces and uniformly continuous maps! See Isbell's "Uniform spaces" (1964), Exercise III.8(c). In fact, the product of any (possibly infinite) collection of quotient maps is a quotient map in the uniform category. (Quotient maps are defined in any concrete category over the category of sets, as explained e.g. in The Joy of Cats.)

Added still later: With this in mind, I no longer see a good reason for myself to think at all about the quotient topology on $\pi_1(X)$. As you can see for instance from my comment here, I have good reasons to believe that quotient topology is a sensible notion only when it coincides with the topology of quotient uniformity (i.e. for quotients of compact spaces), and otherwise the topology of quotient uniformity is the way to go. The present situation does not seem to be an exception.

Still more: While the relation of the two topologies on $\pi_1$ might be not so easy to comprehend, the topology of the quotient uniformity on $\pi_1$ relates in a seemingly more comprehensible way directly to the topologized Steenrod $\pi_1$ (whose relation to the fundamental pro-group is clear). The Steenrod $\pi_1(X)$ can be defined as $\pi_1(\holim P_i)$, where the compact metrizable space $X$ is the inverse limit of the polyhedra $P_i$ and $P_0=pt$. (If $P_0\ne pt$ we can shift the indices by one and add a new $P_0$, namely $pt$.) If you think about the construction of $\holim$, you will see that $\pi_1(\holim P_i)$ is the set of equivalence classes of level-preserving base-ray-preserving maps $S^1\times [0,\infty)\to P_{[0,\infty)}$ under the relation of level-preserving base-ray-preserving homotopy, where $P_{[0,\infty)}$ denotes the mapping telescope of the inverse sequence $\dots\to P_1\to P_0$ (glued out of the mapping cylinders $P_{[i,i+1]}$ of the individual bonding maps). Since $P_0=pt$, "level-preserving" can in fact be replaced by "proper" (see Lemma 2.5 in "Steenrod homotopy"). In other words, the Steenrod $\pi_1(X)$ is the set of equivalence classes of base-ray preserving maps from $D^2$ to the one-point compactification $P_{[0,\infty)}^+$ of the mapping telescope such that the preimage of the point at infinity is precisely the center of the disk (and nothing else).

Now it is not hard to see that the topology on the Steenrod $\pi_1(X)$ (induced from the inverse limit topology on the Cech $\pi_1(X)$) is precisely the topology of the quotient uniformity. The uniformity is on a subspace of the space of all continuous (=uniformly continuous) maps from $D^2$ to $P_{[0,\infty)}^+$, and so it is the subspace uniformity (of the uniformity of uniform convergence).

The continuous map from $\pi_1(X)$ with the topology of the quotient uniformity to the Steenrod $\pi_1(X)$ is given by Milnor's lemma (see Lemma 2.1 in Steenrod homotopy): every map $S^1\to X$ extends to a level-preserving map $S^1\x [0,\infty]\to P_{[0,\infty]}$, where $P_{[0,\infty]}$ is Milnor's compactification of the mapping telescope by a copy of $X$, and this extension is well-defined up to homotopy through such extensions. Also close maps have close extensions (from the proof of Milnor's lemma). The quotient $P_{[0,\infty]}/X$ is homeomorphic to $P_{[0,\infty]}^+$, so the close extensions descend to close representatives of elements of the Steenrod $\pi_1$.

Still more: While the relation of the two topologies on $\pi_1$ might be not so easy to comprehend, the topology of the quotient uniformity on $\pi_1$ relates in a seemingly more comprehensible way directly to the topologized Steenrod $\pi_1$ (whose relation to the fundamental pro-group is clear). The Steenrod $\pi_1(X)$ can be defined as $\pi_1(holim P_i)$, where the compact metrizable space $X$ is the inverse limit of the polyhedra $P_i$ and $P_0=pt$. (If $P_0\ne pt$ we can shift the indices by one and add a new $P_0$, namely $pt$.) If you think about the construction of $holim$, you will see that $\pi_1(holim P_i)$ is the set of equivalence classes of level-preserving base-ray-preserving maps $S^1\times [0,\infty)\to P_{[0,\infty)}$ under the relation of level-preserving base-ray-preserving homotopy, where $P_{[0,\infty)}$ denotes the mapping telescope of the inverse sequence $\dots\to P_1\to P_0$ (glued out of the mapping cylinders $P_{[i,i+1]}$ of the individual bonding maps). Since $P_0=pt$, "level-preserving" can in fact be replaced by "proper" (see Lemma 2.5 in "Steenrod homotopy"). In other words, the Steenrod $\pi_1(X)$ is the set of equivalence classes of base-ray preserving maps from $D^2$ to the one-point compactification $P_{[0,\infty)}^+$ of the mapping telescope such that the preimage of the point at infinity is precisely the center of the disk (and nothing else).

Now it is not hard to see that the topology on the Steenrod $\pi_1(X)$ (induced from the inverse limit topology on the Cech $\pi_1(X)$ ) is precisely the topology of the quotient uniformity. The uniformity is on a subspace of the space of all continuous (=uniformly continuous) maps from $D^2$ to $P_{[0,\infty)}^+$, and so it is the subspace uniformity (of the uniformity of uniform convergence).

The continuous map from $\pi_1(X)$ with the topology of the quotient uniformity to the Steenrod $\pi_1(X)$ is given by Milnor's lemma (see Lemma 2.1 in Steenrod homotopy): every map $S^1\to X$ extends to a level-preserving map $S^1\times [0,\infty]\to P_{[0,\infty]}$, where $P_{[0,\infty]}$ is Milnor's compactification of the mapping telescope by a copy of $X$, and this extension is well-defined up to homotopy through such extensions. Also close maps have close extensions (from the proof of Milnor's lemma). The quotient $P_{[0,\infty]}/X$ is homeomorphic to $P_{[0,\infty]}^+$, so the close extensions descend to close representatives of elements of the Steenrod $\pi_1$.