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My exercise is:

Let $F_n$ be the free group on $n$ generators. Determine an infinite succession of topological spaces $X_1$, $X_2$, $\ldots$ connected by arcs and such that the following properties are valid:

  • $X_i$ is not homeomorphic to $X_j$ for $i \neq j$;
  • the fundamental group of $X_i$ is equal to $F_n$ for every $i$ and for every $n$.

I plan to consider $X_i={}$ ($S_1 \wedge S_1 \wedge \ldots$ infinite times) united with infinite points. For example, $X_1=(S_1) \cup$ one point, $X_2=(S_1 \wedge S_1) \cup$ two points, et cetera. This space that I considered, is it connected by arcs? And does this space satisfy those two properties?

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    $\begingroup$ Do you mean $S^1\vee S^1$ (with a \vee) or $S^1\wedge S^1$ (with a \wedge)? $\endgroup$ – Tyrone Mar 4 at 10:04
  • $\begingroup$ Also, what do you mean "the fundamental group of $X_i$ is equal to $F_n$ for every $i$ and every $n$"? And note that unless it is trivial, the fundamental group of any space $X$ is never equal to an abstract group. Rather it is isomorphic to that group. $\endgroup$ – Tyrone Mar 4 at 10:07
  • $\begingroup$ @Tyrone How do you mean? The fundamental group of the circle fails to be $\mathbb{Z}$ in the same way the fundamental group of the disk fails to be the trivial group for almost any choice you make to represent the trivial group. Both consist of homotopy classes of functions. $\endgroup$ – Connor Malin Mar 4 at 10:12
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    $\begingroup$ @ConnorMalin, I'm certainly not someone to discuss foundational topics with, so feel free to provide a correction if you think my statement is wrong. Since any 1-element set uniquely becomes the trivial group with no possible choices to make, really I'm assuming a universe in which all 1-element sets are considered equal. Perhaps to be safe, always write isomorphic. $\endgroup$ – Tyrone Mar 4 at 10:23
  • $\begingroup$ I'm having trouble parsing this condition: "the fundamental group of $𝑋_𝑖$ is equal to $𝐹_𝑛$ for every $𝑖$ and for every $𝑛$." The way its phrased suggests that for EACH $i$ the fixed space $X_i$ is supposed to have $\pi_1(X_i)\cong F_n$ for EVERY $n$, which doesn't make sense since these groups are not isomorphic. $\endgroup$ – William Mar 4 at 13:23

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