Proving a space with trivial topology is contractil
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Here is the quesion: Let be $X$ a topological space with trivial topology $mathcal{T} = {emptyset,X}$. Prove $X$ is contractil (any constant map $f(x) = x_0$ is homotopical to the identity map $Id$).
WAY #1: Define $F : X times [0,1] to X$ as $F(x,0) := x$ and $F(x,t) := x_0$ for $t in (0,1]$. $F$ is continous (cause $mathcal{T}$ is trivial) and satisfies $F(x_0,t) = x_0$ (is it necessary this one?), $F(x,0) = Id$ and $F(x,1) = x_0,$ so it is an homotopy.
WAY #2: We define $F : X times [0,1] to X$ as $F(x,t) := x_0t + (1-t)x$. Again, $F$ is continous (cause the topology of $X$) and satisfies $F(x,0) = Id$, $F(x,1) = x_0,$ so it is an homotopy.
What way is right?
Thank you all.
algebraic-topology
asked Dec 16 '18 at 18:44
LH8LH8
1438
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add a comment |
$begingroup$
Here is the quesion: Let be $X$ a topological space with trivial topology $mathcal{T} = {emptyset,X}$. Prove $X$ is contractil (any constant map $f(x) = x_0$ is homotopical to the identity map $Id$).
WAY #1: Define $F : X times [0,1] to X$ as $F(x,0) := x$ and $F(x,t) := x_0$ for $t in (0,1]$. $F$ is continous (cause $mathcal{T}$ is trivial) and satisfies $F(x_0,t) = x_0$ (is it necessary this one?), $F(x,0) = Id$ and $F(x,1) = x_0,$ so it is an homotopy.
WAY #2: We define $F : X times [0,1] to X$ as $F(x,t) := x_0t + (1-t)x$. Again, $F$ is continous (cause the topology of $X$) and satisfies $F(x,0) = Id$, $F(x,1) = x_0,$ so it is an homotopy.
What way is right?
Thank you all.
algebraic-topology
asked Dec 16 '18 at 18:44
LH8LH8
1438
$endgroup$
add a comment |
$begingroup$
Here is the quesion: Let be $X$ a topological space with trivial topology $mathcal{T} = {emptyset,X}$. Prove $X$ is contractil (any constant map $f(x) = x_0$ is homotopical to the identity map $Id$).
WAY #1: Define $F : X times [0,1] to X$ as $F(x,0) := x$ and $F(x,t) := x_0$ for $t in (0,1]$. $F$ is continous (cause $mathcal{T}$ is trivial) and satisfies $F(x_0,t) = x_0$ (is it necessary this one?), $F(x,0) = Id$ and $F(x,1) = x_0,$ so it is an homotopy.
WAY #2: We define $F : X times [0,1] to X$ as $F(x,t) := x_0t + (1-t)x$. Again, $F$ is continous (cause the topology of $X$) and satisfies $F(x,0) = Id$, $F(x,1) = x_0,$ so it is an homotopy.
What way is right?
Thank you all.
algebraic-topology
asked Dec 16 '18 at 18:44
LH8LH8
1438
$endgroup$
Here is the quesion: Let be $X$ a topological space with trivial topology $mathcal{T} = {emptyset,X}$. Prove $X$ is contractil (any constant map $f(x) = x_0$ is homotopical to the identity map $Id$).
WAY #1: Define $F : X times [0,1] to X$ as $F(x,0) := x$ and $F(x,t) := x_0$ for $t in (0,1]$. $F$ is continous (cause $mathcal{T}$ is trivial) and satisfies $F(x_0,t) = x_0$ (is it necessary this one?), $F(x,0) = Id$ and $F(x,1) = x_0,$ so it is an homotopy.
WAY #2: We define $F : X times [0,1] to X$ as $F(x,t) := x_0t + (1-t)x$. Again, $F$ is continous (cause the topology of $X$) and satisfies $F(x,0) = Id$, $F(x,1) = x_0,$ so it is an homotopy.
What way is right?
Thank you all.
algebraic-topology
algebraic-topology
asked Dec 16 '18 at 18:44
LH8LH8
1438
asked Dec 16 '18 at 18:44
LH8LH8
1438
asked Dec 16 '18 at 18:44
LH8LH8
1438
asked Dec 16 '18 at 18:44
LH8LH8
1438
asked Dec 16 '18 at 18:44
LH8LH8
1438
1438
add a comment |
add a comment |
1 Answer
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Way #1 is right and way #2 is not. The (is it necessary this one?) was indeed not necessary. Way #2 is not right, because $X$ need not be a vector space, so $x_0t+(1-t)x$ need not be well-defined.
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
$endgroup$
$begingroup$
Thanks, SmileyCraft! :)
$endgroup$
– LH8
Dec 16 '18 at 19:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Way #1 is right and way #2 is not. The (is it necessary this one?) was indeed not necessary. Way #2 is not right, because $X$ need not be a vector space, so $x_0t+(1-t)x$ need not be well-defined.
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
$endgroup$
$begingroup$
Thanks, SmileyCraft! :)
$endgroup$
– LH8
Dec 16 '18 at 19:03
add a comment |
$begingroup$
Way #1 is right and way #2 is not. The (is it necessary this one?) was indeed not necessary. Way #2 is not right, because $X$ need not be a vector space, so $x_0t+(1-t)x$ need not be well-defined.
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
$endgroup$
$begingroup$
Thanks, SmileyCraft! :)
$endgroup$
– LH8
Dec 16 '18 at 19:03
add a comment |
$begingroup$
Way #1 is right and way #2 is not. The (is it necessary this one?) was indeed not necessary. Way #2 is not right, because $X$ need not be a vector space, so $x_0t+(1-t)x$ need not be well-defined.
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
$endgroup$
Way #1 is right and way #2 is not. The (is it necessary this one?) was indeed not necessary. Way #2 is not right, because $X$ need not be a vector space, so $x_0t+(1-t)x$ need not be well-defined.
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
answered Dec 16 '18 at 18:50
SmileyCraftSmileyCraft
3,626519
3,626519
$begingroup$
Thanks, SmileyCraft! :)
$endgroup$
– LH8
Dec 16 '18 at 19:03
add a comment |
$begingroup$
Thanks, SmileyCraft! :)
$endgroup$
– LH8
Dec 16 '18 at 19:03
$begingroup$
Thanks, SmileyCraft! :)
$endgroup$
– LH8
Dec 16 '18 at 19:03
$begingroup$
Thanks, SmileyCraft! :)
$endgroup$
– LH8
Dec 16 '18 at 19:03
add a comment |
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