Mixing
A connected, but not path-connected, space whose fundamental group depends on the basepoint
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It is a well known result that the fundamental group of a path-connected space is independent (up to isomorphism) of the choice of the basepoint.
Can someone provide an explicit example of a connected, but not path-connected, space for which the fundamental group does indeed depend on the basepoint?
general-topology algebraic-topology examples-counterexamples
asked Jan 15 at 20:59
la flacala flaca
1,602518
$endgroup$
add a comment |
$begingroup$
It is a well known result that the fundamental group of a path-connected space is independent (up to isomorphism) of the choice of the basepoint.
Can someone provide an explicit example of a connected, but not path-connected, space for which the fundamental group does indeed depend on the basepoint?
general-topology algebraic-topology examples-counterexamples
asked Jan 15 at 20:59
la flacala flaca
1,602518
$endgroup$
6
$begingroup$
How about sticking a loop onto the end of the topologist's sine curve?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 21:02
2
$begingroup$
The fundamental group is entirely about maps from paths into your space, so it would be ridiculous if it could detect things in different path-components. Generalizing Lord Shark's example, call the "path disconnected wedge" (a name I made up for an operation I made up) of two pointed spaces $X$ and $Y$ to be what you get when you take the topologist's sine curve (with the arc at the end), reflecting it across the $y$-axis, and gluing the two spaces to the two 'endpoints' of this connected but not path-connected, 'arc'. Then the f.g. of the $X$ component is $pi_1 X$, and similarly with $Y$.
$endgroup$
– Mike Miller
Jan 15 at 21:19
add a comment |
$begingroup$
It is a well known result that the fundamental group of a path-connected space is independent (up to isomorphism) of the choice of the basepoint.
Can someone provide an explicit example of a connected, but not path-connected, space for which the fundamental group does indeed depend on the basepoint?
general-topology algebraic-topology examples-counterexamples
asked Jan 15 at 20:59
la flacala flaca
1,602518
$endgroup$
It is a well known result that the fundamental group of a path-connected space is independent (up to isomorphism) of the choice of the basepoint.
Can someone provide an explicit example of a connected, but not path-connected, space for which the fundamental group does indeed depend on the basepoint?
general-topology algebraic-topology examples-counterexamples
general-topology algebraic-topology examples-counterexamples
asked Jan 15 at 20:59
la flacala flaca
1,602518
asked Jan 15 at 20:59
la flacala flaca
1,602518
asked Jan 15 at 20:59
la flacala flaca
1,602518
asked Jan 15 at 20:59
la flacala flaca
1,602518
asked Jan 15 at 20:59
la flacala flaca
1,602518
1,602518
6
$begingroup$
How about sticking a loop onto the end of the topologist's sine curve?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 21:02
2
$begingroup$
The fundamental group is entirely about maps from paths into your space, so it would be ridiculous if it could detect things in different path-components. Generalizing Lord Shark's example, call the "path disconnected wedge" (a name I made up for an operation I made up) of two pointed spaces $X$ and $Y$ to be what you get when you take the topologist's sine curve (with the arc at the end), reflecting it across the $y$-axis, and gluing the two spaces to the two 'endpoints' of this connected but not path-connected, 'arc'. Then the f.g. of the $X$ component is $pi_1 X$, and similarly with $Y$.
$endgroup$
– Mike Miller
Jan 15 at 21:19
add a comment |
6
$begingroup$
How about sticking a loop onto the end of the topologist's sine curve?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 21:02
2
$begingroup$
The fundamental group is entirely about maps from paths into your space, so it would be ridiculous if it could detect things in different path-components. Generalizing Lord Shark's example, call the "path disconnected wedge" (a name I made up for an operation I made up) of two pointed spaces $X$ and $Y$ to be what you get when you take the topologist's sine curve (with the arc at the end), reflecting it across the $y$-axis, and gluing the two spaces to the two 'endpoints' of this connected but not path-connected, 'arc'. Then the f.g. of the $X$ component is $pi_1 X$, and similarly with $Y$.
$endgroup$
– Mike Miller
Jan 15 at 21:19
6
6
$begingroup$
How about sticking a loop onto the end of the topologist's sine curve?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 21:02
$begingroup$
How about sticking a loop onto the end of the topologist's sine curve?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 21:02
2
2
$begingroup$
The fundamental group is entirely about maps from paths into your space, so it would be ridiculous if it could detect things in different path-components. Generalizing Lord Shark's example, call the "path disconnected wedge" (a name I made up for an operation I made up) of two pointed spaces $X$ and $Y$ to be what you get when you take the topologist's sine curve (with the arc at the end), reflecting it across the $y$-axis, and gluing the two spaces to the two 'endpoints' of this connected but not path-connected, 'arc'. Then the f.g. of the $X$ component is $pi_1 X$, and similarly with $Y$.
$endgroup$
– Mike Miller
Jan 15 at 21:19
$begingroup$
The fundamental group is entirely about maps from paths into your space, so it would be ridiculous if it could detect things in different path-components. Generalizing Lord Shark's example, call the "path disconnected wedge" (a name I made up for an operation I made up) of two pointed spaces $X$ and $Y$ to be what you get when you take the topologist's sine curve (with the arc at the end), reflecting it across the $y$-axis, and gluing the two spaces to the two 'endpoints' of this connected but not path-connected, 'arc'. Then the f.g. of the $X$ component is $pi_1 X$, and similarly with $Y$.
$endgroup$
– Mike Miller
Jan 15 at 21:19
add a comment |
1 Answer
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Let $X = {(sin(t), cos(t), arctan(t)) in mathbb{R}^3 mid t in mathbb{R}}$, then let $Y = S^1 times {-pi/4}$ and $Z = (S^1 times {pi/4})vee W$ for any path-connected space $W$ with non-trivial fundamental group.
Now take the space $X cup Y cup Z$ which is connected ($X$ gets arbitrarily close to both $Y$ and $Z$ and so cannot be separated from either by open neighbourhoods), but has three distinct path-components, $X$, $Y$ and $Z$, with $pi_1(X) = 1$, $pi_1(Y) = mathbb{Z}$ and $pi_1(Z) =mathbb{Z} ast pi_1(W)$ which are each non-isomorphic.
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
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1 Answer
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1 Answer
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$begingroup$
Let $X = {(sin(t), cos(t), arctan(t)) in mathbb{R}^3 mid t in mathbb{R}}$, then let $Y = S^1 times {-pi/4}$ and $Z = (S^1 times {pi/4})vee W$ for any path-connected space $W$ with non-trivial fundamental group.
Now take the space $X cup Y cup Z$ which is connected ($X$ gets arbitrarily close to both $Y$ and $Z$ and so cannot be separated from either by open neighbourhoods), but has three distinct path-components, $X$, $Y$ and $Z$, with $pi_1(X) = 1$, $pi_1(Y) = mathbb{Z}$ and $pi_1(Z) =mathbb{Z} ast pi_1(W)$ which are each non-isomorphic.
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
$endgroup$
add a comment |
$begingroup$
Let $X = {(sin(t), cos(t), arctan(t)) in mathbb{R}^3 mid t in mathbb{R}}$, then let $Y = S^1 times {-pi/4}$ and $Z = (S^1 times {pi/4})vee W$ for any path-connected space $W$ with non-trivial fundamental group.
Now take the space $X cup Y cup Z$ which is connected ($X$ gets arbitrarily close to both $Y$ and $Z$ and so cannot be separated from either by open neighbourhoods), but has three distinct path-components, $X$, $Y$ and $Z$, with $pi_1(X) = 1$, $pi_1(Y) = mathbb{Z}$ and $pi_1(Z) =mathbb{Z} ast pi_1(W)$ which are each non-isomorphic.
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
$endgroup$
add a comment |
$begingroup$
Let $X = {(sin(t), cos(t), arctan(t)) in mathbb{R}^3 mid t in mathbb{R}}$, then let $Y = S^1 times {-pi/4}$ and $Z = (S^1 times {pi/4})vee W$ for any path-connected space $W$ with non-trivial fundamental group.
Now take the space $X cup Y cup Z$ which is connected ($X$ gets arbitrarily close to both $Y$ and $Z$ and so cannot be separated from either by open neighbourhoods), but has three distinct path-components, $X$, $Y$ and $Z$, with $pi_1(X) = 1$, $pi_1(Y) = mathbb{Z}$ and $pi_1(Z) =mathbb{Z} ast pi_1(W)$ which are each non-isomorphic.
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
$endgroup$
Let $X = {(sin(t), cos(t), arctan(t)) in mathbb{R}^3 mid t in mathbb{R}}$, then let $Y = S^1 times {-pi/4}$ and $Z = (S^1 times {pi/4})vee W$ for any path-connected space $W$ with non-trivial fundamental group.
Now take the space $X cup Y cup Z$ which is connected ($X$ gets arbitrarily close to both $Y$ and $Z$ and so cannot be separated from either by open neighbourhoods), but has three distinct path-components, $X$, $Y$ and $Z$, with $pi_1(X) = 1$, $pi_1(Y) = mathbb{Z}$ and $pi_1(Z) =mathbb{Z} ast pi_1(W)$ which are each non-isomorphic.
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
answered Jan 16 at 13:03
Dan RustDan Rust
22.8k114884
22.8k114884
add a comment |
add a comment |
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6
$begingroup$
How about sticking a loop onto the end of the topologist's sine curve?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 21:02
2
$begingroup$
The fundamental group is entirely about maps from paths into your space, so it would be ridiculous if it could detect things in different path-components. Generalizing Lord Shark's example, call the "path disconnected wedge" (a name I made up for an operation I made up) of two pointed spaces $X$ and $Y$ to be what you get when you take the topologist's sine curve (with the arc at the end), reflecting it across the $y$-axis, and gluing the two spaces to the two 'endpoints' of this connected but not path-connected, 'arc'. Then the f.g. of the $X$ component is $pi_1 X$, and similarly with $Y$.
$endgroup$
– Mike Miller
Jan 15 at 21:19