Prove that there exists a long exact sequence…
$begingroup$
Let $f, g : X to Y$ be two continuous maps. Consider the space $Z$ which is obtained from the disjoint union of $Y$ with $(X times [0, 1])$ by identifying $(x, 0) sim f (x), (x, 1) sim g(x),$ for each $x$ in $X$. Let $i$ denote the natural inclusion of $Y$ into $Z$. Prove that there exists a long exact sequence of the following form:
$$cdotslongrightarrow H_n(X) longrightarrow H_n(Y ) xrightarrow{;i_*;} H_n(Z) longrightarrow H_{n−1}(X) longrightarrowcdots$$
algebraic-topology
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
$endgroup$
add a comment |
$begingroup$
Let $f, g : X to Y$ be two continuous maps. Consider the space $Z$ which is obtained from the disjoint union of $Y$ with $(X times [0, 1])$ by identifying $(x, 0) sim f (x), (x, 1) sim g(x),$ for each $x$ in $X$. Let $i$ denote the natural inclusion of $Y$ into $Z$. Prove that there exists a long exact sequence of the following form:
$$cdotslongrightarrow H_n(X) longrightarrow H_n(Y ) xrightarrow{;i_*;} H_n(Z) longrightarrow H_{n−1}(X) longrightarrowcdots$$
algebraic-topology
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
$endgroup$
add a comment |
$begingroup$
Let $f, g : X to Y$ be two continuous maps. Consider the space $Z$ which is obtained from the disjoint union of $Y$ with $(X times [0, 1])$ by identifying $(x, 0) sim f (x), (x, 1) sim g(x),$ for each $x$ in $X$. Let $i$ denote the natural inclusion of $Y$ into $Z$. Prove that there exists a long exact sequence of the following form:
$$cdotslongrightarrow H_n(X) longrightarrow H_n(Y ) xrightarrow{;i_*;} H_n(Z) longrightarrow H_{n−1}(X) longrightarrowcdots$$
algebraic-topology
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
$endgroup$
Let $f, g : X to Y$ be two continuous maps. Consider the space $Z$ which is obtained from the disjoint union of $Y$ with $(X times [0, 1])$ by identifying $(x, 0) sim f (x), (x, 1) sim g(x),$ for each $x$ in $X$. Let $i$ denote the natural inclusion of $Y$ into $Z$. Prove that there exists a long exact sequence of the following form:
$$cdotslongrightarrow H_n(X) longrightarrow H_n(Y ) xrightarrow{;i_*;} H_n(Z) longrightarrow H_{n−1}(X) longrightarrowcdots$$
algebraic-topology
algebraic-topology
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
edited Jul 17 '13 at 4:33
Zev Chonoles
110k16228426
110k16228426
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
asked Jul 17 '13 at 4:29
dunkindonutsdunkindonuts
684
684
add a comment |
add a comment |
1 Answer
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$begingroup$
The space $Z$ you construct is also called the double mapping cylinder $M(f,g)$. By cutting it in the middle, you can see it as the union of two mapping cylinders $M(f),M(g)$ identified at the ends $X times {0}$.
Above is a simple rough (!) sketch. Now apply the Mayer-Vietoris sequence to this union.
A good example of this is to take the two maps $f,g: S^1 to S^1$ given by $zmapsto z^2, z mapsto z^3$ respectively. The coequaliser of $f,g$, obtained by identifying $f(z) sim g(z)$, is not even Hausdorff. But $M(f,g)$ is a nice CW-complex.
The fundamental group of this example is also of interest. It has two generators say $a,b$ and one relation $a^2=b^3$.
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The space $Z$ you construct is also called the double mapping cylinder $M(f,g)$. By cutting it in the middle, you can see it as the union of two mapping cylinders $M(f),M(g)$ identified at the ends $X times {0}$.
Above is a simple rough (!) sketch. Now apply the Mayer-Vietoris sequence to this union.
A good example of this is to take the two maps $f,g: S^1 to S^1$ given by $zmapsto z^2, z mapsto z^3$ respectively. The coequaliser of $f,g$, obtained by identifying $f(z) sim g(z)$, is not even Hausdorff. But $M(f,g)$ is a nice CW-complex.
The fundamental group of this example is also of interest. It has two generators say $a,b$ and one relation $a^2=b^3$.
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
$endgroup$
add a comment |
$begingroup$
The space $Z$ you construct is also called the double mapping cylinder $M(f,g)$. By cutting it in the middle, you can see it as the union of two mapping cylinders $M(f),M(g)$ identified at the ends $X times {0}$.
Above is a simple rough (!) sketch. Now apply the Mayer-Vietoris sequence to this union.
A good example of this is to take the two maps $f,g: S^1 to S^1$ given by $zmapsto z^2, z mapsto z^3$ respectively. The coequaliser of $f,g$, obtained by identifying $f(z) sim g(z)$, is not even Hausdorff. But $M(f,g)$ is a nice CW-complex.
The fundamental group of this example is also of interest. It has two generators say $a,b$ and one relation $a^2=b^3$.
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
$endgroup$
add a comment |
$begingroup$
The space $Z$ you construct is also called the double mapping cylinder $M(f,g)$. By cutting it in the middle, you can see it as the union of two mapping cylinders $M(f),M(g)$ identified at the ends $X times {0}$.
Above is a simple rough (!) sketch. Now apply the Mayer-Vietoris sequence to this union.
A good example of this is to take the two maps $f,g: S^1 to S^1$ given by $zmapsto z^2, z mapsto z^3$ respectively. The coequaliser of $f,g$, obtained by identifying $f(z) sim g(z)$, is not even Hausdorff. But $M(f,g)$ is a nice CW-complex.
The fundamental group of this example is also of interest. It has two generators say $a,b$ and one relation $a^2=b^3$.
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
$endgroup$
The space $Z$ you construct is also called the double mapping cylinder $M(f,g)$. By cutting it in the middle, you can see it as the union of two mapping cylinders $M(f),M(g)$ identified at the ends $X times {0}$.
Above is a simple rough (!) sketch. Now apply the Mayer-Vietoris sequence to this union.
A good example of this is to take the two maps $f,g: S^1 to S^1$ given by $zmapsto z^2, z mapsto z^3$ respectively. The coequaliser of $f,g$, obtained by identifying $f(z) sim g(z)$, is not even Hausdorff. But $M(f,g)$ is a nice CW-complex.
The fundamental group of this example is also of interest. It has two generators say $a,b$ and one relation $a^2=b^3$.
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
edited Dec 23 '18 at 21:13
Glorfindel
3,41981830
3,41981830
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
answered Jul 17 '13 at 9:54
Ronnie BrownRonnie Brown
12k12938
12k12938
add a comment |
add a comment |
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