Mixing
Homotopy invariance of relative homology
up vote
3
down vote
favorite
I have studied the proof of the homotopy invariance of singular homology. If I understood correctly, the key step is to transform a homotopy $F : X times I to Y$ between the maps $f, g : X to Y$, into a chain homotopy between the induced chain maps $f_n, g_n : X_n to Y_n$. For this purpose, we define the prism natural transformation $P_n : mathcal C_n to mathcal D_n$ between the functors $mathcal C_n(X) = X_n$ and $mathcal D_n(X) = (X times I)_{n+1}$. This allows us to form the following chain homotopy:
Now I am trying to obtain a similar result for relative homology. Let $A subset X$ and $B subset Y$ be subspaces, and let $F : (X times I, A times I) to (Y,B)$ be a homotopy between $f, g : (X,A) to (Y,B)$. I want to construct the following commutative parallelepiped:
The homotopy invariance of absolute homology gives me the left and middle faces. If two chains in $X_n$ differ by a chain in $A_n$, their prisms differ by a chain in $(A times I)_{n+1}$, so the right face is well-defined as well. This gives me a copy of the first diagram, with $X_n$, $Y_n$, $(X times I)_n$ replaced with $X_n / A_n$, $Y_n / B_n$, $(X times I)_n / (A times I)_n$ respectively. The diagonals
$$frac {X_n} {A_n} longrightarrow frac {(X times I)_{n+1}} {(A times I)_{n+1}} longrightarrow frac {Y_{n+1}} {B_{n+1}}$$
constitute a chain homotopy between $f$ and $g$, so the relative homology groups are equal. Does this work? Is there anything missing?
algebraic-topology homology-cohomology
asked Nov 18 at 1:25
pyon
26719
add a comment |
up vote
3
down vote
favorite
I have studied the proof of the homotopy invariance of singular homology. If I understood correctly, the key step is to transform a homotopy $F : X times I to Y$ between the maps $f, g : X to Y$, into a chain homotopy between the induced chain maps $f_n, g_n : X_n to Y_n$. For this purpose, we define the prism natural transformation $P_n : mathcal C_n to mathcal D_n$ between the functors $mathcal C_n(X) = X_n$ and $mathcal D_n(X) = (X times I)_{n+1}$. This allows us to form the following chain homotopy:
Now I am trying to obtain a similar result for relative homology. Let $A subset X$ and $B subset Y$ be subspaces, and let $F : (X times I, A times I) to (Y,B)$ be a homotopy between $f, g : (X,A) to (Y,B)$. I want to construct the following commutative parallelepiped:
The homotopy invariance of absolute homology gives me the left and middle faces. If two chains in $X_n$ differ by a chain in $A_n$, their prisms differ by a chain in $(A times I)_{n+1}$, so the right face is well-defined as well. This gives me a copy of the first diagram, with $X_n$, $Y_n$, $(X times I)_n$ replaced with $X_n / A_n$, $Y_n / B_n$, $(X times I)_n / (A times I)_n$ respectively. The diagonals
$$frac {X_n} {A_n} longrightarrow frac {(X times I)_{n+1}} {(A times I)_{n+1}} longrightarrow frac {Y_{n+1}} {B_{n+1}}$$
constitute a chain homotopy between $f$ and $g$, so the relative homology groups are equal. Does this work? Is there anything missing?
algebraic-topology homology-cohomology
asked Nov 18 at 1:25
pyon
26719
4
The prism operator maps $ mathcal C_n(A)$ to $mathcal D_n(B)$ so it induces an operator $: mathcal C_n(X,A) to mathcal D_n(Y,B)$ on the relative chain groups. Then the proof goes through as in the absolute case. Or am I missing something?
– Matematleta
Nov 18 at 4:15
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have studied the proof of the homotopy invariance of singular homology. If I understood correctly, the key step is to transform a homotopy $F : X times I to Y$ between the maps $f, g : X to Y$, into a chain homotopy between the induced chain maps $f_n, g_n : X_n to Y_n$. For this purpose, we define the prism natural transformation $P_n : mathcal C_n to mathcal D_n$ between the functors $mathcal C_n(X) = X_n$ and $mathcal D_n(X) = (X times I)_{n+1}$. This allows us to form the following chain homotopy:
Now I am trying to obtain a similar result for relative homology. Let $A subset X$ and $B subset Y$ be subspaces, and let $F : (X times I, A times I) to (Y,B)$ be a homotopy between $f, g : (X,A) to (Y,B)$. I want to construct the following commutative parallelepiped:
The homotopy invariance of absolute homology gives me the left and middle faces. If two chains in $X_n$ differ by a chain in $A_n$, their prisms differ by a chain in $(A times I)_{n+1}$, so the right face is well-defined as well. This gives me a copy of the first diagram, with $X_n$, $Y_n$, $(X times I)_n$ replaced with $X_n / A_n$, $Y_n / B_n$, $(X times I)_n / (A times I)_n$ respectively. The diagonals
$$frac {X_n} {A_n} longrightarrow frac {(X times I)_{n+1}} {(A times I)_{n+1}} longrightarrow frac {Y_{n+1}} {B_{n+1}}$$
constitute a chain homotopy between $f$ and $g$, so the relative homology groups are equal. Does this work? Is there anything missing?
algebraic-topology homology-cohomology
asked Nov 18 at 1:25
pyon
26719
I have studied the proof of the homotopy invariance of singular homology. If I understood correctly, the key step is to transform a homotopy $F : X times I to Y$ between the maps $f, g : X to Y$, into a chain homotopy between the induced chain maps $f_n, g_n : X_n to Y_n$. For this purpose, we define the prism natural transformation $P_n : mathcal C_n to mathcal D_n$ between the functors $mathcal C_n(X) = X_n$ and $mathcal D_n(X) = (X times I)_{n+1}$. This allows us to form the following chain homotopy:
Now I am trying to obtain a similar result for relative homology. Let $A subset X$ and $B subset Y$ be subspaces, and let $F : (X times I, A times I) to (Y,B)$ be a homotopy between $f, g : (X,A) to (Y,B)$. I want to construct the following commutative parallelepiped:
The homotopy invariance of absolute homology gives me the left and middle faces. If two chains in $X_n$ differ by a chain in $A_n$, their prisms differ by a chain in $(A times I)_{n+1}$, so the right face is well-defined as well. This gives me a copy of the first diagram, with $X_n$, $Y_n$, $(X times I)_n$ replaced with $X_n / A_n$, $Y_n / B_n$, $(X times I)_n / (A times I)_n$ respectively. The diagonals
$$frac {X_n} {A_n} longrightarrow frac {(X times I)_{n+1}} {(A times I)_{n+1}} longrightarrow frac {Y_{n+1}} {B_{n+1}}$$
constitute a chain homotopy between $f$ and $g$, so the relative homology groups are equal. Does this work? Is there anything missing?
algebraic-topology homology-cohomology
algebraic-topology homology-cohomology
asked Nov 18 at 1:25
pyon
26719
asked Nov 18 at 1:25
pyon
26719
edited 2 days ago
asked Nov 18 at 1:25
pyon
26719
asked Nov 18 at 1:25
pyon
26719
asked Nov 18 at 1:25
pyon
26719
26719
4
The prism operator maps $ mathcal C_n(A)$ to $mathcal D_n(B)$ so it induces an operator $: mathcal C_n(X,A) to mathcal D_n(Y,B)$ on the relative chain groups. Then the proof goes through as in the absolute case. Or am I missing something?
– Matematleta
Nov 18 at 4:15
add a comment |
4
The prism operator maps $ mathcal C_n(A)$ to $mathcal D_n(B)$ so it induces an operator $: mathcal C_n(X,A) to mathcal D_n(Y,B)$ on the relative chain groups. Then the proof goes through as in the absolute case. Or am I missing something?
– Matematleta
Nov 18 at 4:15
4
4
The prism operator maps $ mathcal C_n(A)$ to $mathcal D_n(B)$ so it induces an operator $: mathcal C_n(X,A) to mathcal D_n(Y,B)$ on the relative chain groups. Then the proof goes through as in the absolute case. Or am I missing something?
– Matematleta
Nov 18 at 4:15
The prism operator maps $ mathcal C_n(A)$ to $mathcal D_n(B)$ so it induces an operator $: mathcal C_n(X,A) to mathcal D_n(Y,B)$ on the relative chain groups. Then the proof goes through as in the absolute case. Or am I missing something?
– Matematleta
Nov 18 at 4:15
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The missing step is the actual calculation. By the homotopy invariance of absolute homology, we have
$$partial_{n+1} F_{n+1} P_n gamma + F_n P_{n-1} partial_n gamma = g_n gamma - f_n gamma$$
Let $gamma in X_n$ be a relative cycle. In other words, suppose that $partial_n gamma in A_{n-1}$. Then $F_n P_{n-1} partial_n gamma in B_n$, hence $g_n gamma - f_n gamma$ is a relative boundary, hence $f$ and $g$ have the same relative homology at the level $n$.
answered yesterday
pyon
26719
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The missing step is the actual calculation. By the homotopy invariance of absolute homology, we have
$$partial_{n+1} F_{n+1} P_n gamma + F_n P_{n-1} partial_n gamma = g_n gamma - f_n gamma$$
Let $gamma in X_n$ be a relative cycle. In other words, suppose that $partial_n gamma in A_{n-1}$. Then $F_n P_{n-1} partial_n gamma in B_n$, hence $g_n gamma - f_n gamma$ is a relative boundary, hence $f$ and $g$ have the same relative homology at the level $n$.
answered yesterday
pyon
26719
add a comment |
up vote
0
down vote
accepted
The missing step is the actual calculation. By the homotopy invariance of absolute homology, we have
$$partial_{n+1} F_{n+1} P_n gamma + F_n P_{n-1} partial_n gamma = g_n gamma - f_n gamma$$
Let $gamma in X_n$ be a relative cycle. In other words, suppose that $partial_n gamma in A_{n-1}$. Then $F_n P_{n-1} partial_n gamma in B_n$, hence $g_n gamma - f_n gamma$ is a relative boundary, hence $f$ and $g$ have the same relative homology at the level $n$.
answered yesterday
pyon
26719
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The missing step is the actual calculation. By the homotopy invariance of absolute homology, we have
$$partial_{n+1} F_{n+1} P_n gamma + F_n P_{n-1} partial_n gamma = g_n gamma - f_n gamma$$
Let $gamma in X_n$ be a relative cycle. In other words, suppose that $partial_n gamma in A_{n-1}$. Then $F_n P_{n-1} partial_n gamma in B_n$, hence $g_n gamma - f_n gamma$ is a relative boundary, hence $f$ and $g$ have the same relative homology at the level $n$.
answered yesterday
pyon
26719
The missing step is the actual calculation. By the homotopy invariance of absolute homology, we have
$$partial_{n+1} F_{n+1} P_n gamma + F_n P_{n-1} partial_n gamma = g_n gamma - f_n gamma$$
Let $gamma in X_n$ be a relative cycle. In other words, suppose that $partial_n gamma in A_{n-1}$. Then $F_n P_{n-1} partial_n gamma in B_n$, hence $g_n gamma - f_n gamma$ is a relative boundary, hence $f$ and $g$ have the same relative homology at the level $n$.
answered yesterday
pyon
26719
edited yesterday
answered yesterday
pyon
26719
answered yesterday
pyon
26719
answered yesterday
pyon
26719
26719
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003042%2fhomotopy-invariance-of-relative-homology%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
The prism operator maps $ mathcal C_n(A)$ to $mathcal D_n(B)$ so it induces an operator $: mathcal C_n(X,A) to mathcal D_n(Y,B)$ on the relative chain groups. Then the proof goes through as in the absolute case. Or am I missing something?
– Matematleta
Nov 18 at 4:15