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Related questions: Here and here my use of the word "hole" was criticized. So here I used the word "geometry", and I was informed that I am not considering all the geometry or even the topology, and that I was merely looking at the rational(or real) homotopy types. I do not understand what are rational or real homotopy types except that to make sense of those comments they should better include all the de Rham co(homology), singular cohomology and homotopy group information of the space. But whatever they are, it is interesting to know whether they capture the following type of "hole".

Moreover, one must understand where one gets stuck, when one tries to study pathological spaces. It helps one in understanding where exactly all the extra nice conditions are used, and hopefully this type of approach will help in minimizing the number of false beliefs we unconsciously have.

The line with the double origin is the following space. In the union $\{0\} \times \mathbb R \cup \{ 1 \} \times \mathbb R$, impose the equivalence relation $(0, x) \sim (1, x) $ iff $x \neq 0$.

This space is locally like the real line, ie a $1$-manifold in everything except the Hausdorff condition. It is connected, path-connected and semilocally simply connected. Just the sort of nice space you study in the theory of fundamental group and covering spaces, except for the (significant) pathology that it has one inconvenient extra point violating the Hausdorff-ness.

It seems that the usual methods of computing the fundamental group are not working for this space. Van Kampen's theorem in particular does not apply. Also the covering spaces are weird, just like this space. In fact this space would have been a covering of $\mathbb R$, were it not for the condition that the preimage of every point is a disjoint union of open sets.

So, what if we try to compute the fundamental group of this space? I would be satisfied to know whether it trivial or not. Say, is the collection of homotopy classes of loops based at $1$ nontrivial? It is possible to speculate that a certain loop based at $1$ which passes through both the origins on this special line, in such a way that it passes through the "upper" origin, ie $(1,0)$ on the way left, and it passes through the lower origin, ie $(0,0)$, ought not to be homotopic to the constant loop based at $1$. But how to go about proving/disproving this statement?

In the simplest cases, the fundamental group serves as a measure of the number of 2-dimensional "holes" in a space. It is interesting to know whether they capture the following type of "hole".

This example may look pathological, but one must understand where one gets stuck, when one tries to study pathological spaces. It helps one in understanding where exactly all the extra nice conditions are used, and hopefully this type of approach will help in minimizing the number of false beliefs we unconsciously have.

The line with the double origin is the following space. In the union $\{0\} \times \mathbb R \cup \{ 1 \} \times \mathbb R$, impose the equivalence relation $(0, x) \sim (1, x) $ iff $x \neq 0$.

This space is locally like the real line, ie a $1$-manifold in everything except the Hausdorff condition. It is connected, path-connected and semilocally simply connected. Just the sort of nice space you study in the theory of fundamental group and covering spaces, except for the (significant) pathology that it has one inconvenient extra point violating the Hausdorff-ness.

It seems that the usual methods of computing the fundamental group are not working for this space. Van Kampen's theorem in particular does not apply. Also the covering spaces are weird, just like this space. In fact this space would have been a covering of $\mathbb R$, were it not for the condition that the preimage of every point is a disjoint union of open sets.

So, what if we try to compute the fundamental group of this space? I would be satisfied to know whether it trivial or not. Say, is the collection of homotopy classes of loops based at $1$ nontrivial? It is possible to speculate that a certain loop based at $1$ which passes through both the origins on this special line, in such a way that it passes through the "upper" origin, ie $(1,0)$ on the way left, and it passes through the lower origin, ie $(0,0)$, ought not to be homotopic to the constant loop based at $1$. But how to go about proving/disproving this statement?

What is the fundamental group of the union of $\{0\} \times \mathbb R \cup \{ 1 \} \times \mathbb R$, with the identifications $(0, x) \sim (1, x) $ iff $x \neq 0$?

Related questions: Here and here my use of the word "hole" was criticized. So here I used the word "geometry", and I was informed that I am not considering all the geometry or even the topology, and that I was merely looking at the rational(or real) homotopy types. I do not understand what are rational or real homotopy types except that to make sense of those comments they should better include all the de Rham co(homology), singular cohomology and homotopy group information of the space. But whatever they are, it is interesting to know whether they capture the following type of "hole".

Moreover, one must understand where one gets stuck, when one tries to study pathological spaces. It helps one in understanding where exactly all the extra nice conditions are used, and hopefully this type of approach will help in minimizing the number of false beliefs we unconsciously have.

The line with the double origin is the following space. In the union $\{0\} \times \mathbb R \cup \{ 1 \} \times \mathbb R$, impose the equivalence relation $(0, x) \sim (1, x) $ iff $x \neq 0$.

This space is locally like the real line, ie a $1$-manifold in everything except the Hausdorff condition. It is connected, path-connected and semilocally simply connected. Just the sort of nice space you study in the theory of fundamental group and covering spaces, except for the (significant) pathology that it has one inconvenient extra point violating the Hausdorff-ness.

It seems that the usual methods of computing the fundamental group are not working for this space. Van Kampen's theorem in particular does not apply. Also the covering spaces are weird, just like this space. In fact this space would have been a covering of $\mathbb R$, were it not for the condition that the preimage of every point is a disjoint union of open sets.

So, what if we try to compute the fundamental group of this space? I would be satisfied to know whether it trivial or not. Say, is the collection of homotopy classes of loops based at $1$ nontrivial? It is possible to speculate that a certain loop based at $1$ which passes through both the origins on this special line, in such a way that it passes through the "upper" origin, ie $(1,0)$ on the way left, and it passes through the lower origin, ie $(0,0)$, ought not to be homotopic to the constant loop based at $1$. But how to go about proving/disproving this statement?