Mixing
Formula for relative homotopy groups of products
In Hatcher prop 4.2 he proves that the n-th homotopy group of a product $Xtimes Y$ (for $X$ and $Y$ path-connected) is isomorphic to the product of the n-th homotopy groups of $X$ and $Y$. I wonder if a similar statement is true for relative homotopy groups? I.e. do we have a statement like
$pi_n(Xtimes Y, Atimes B, (x_0,y_0))=pi_n(X,A,x_0)times pi_n(Y,B,y_0)$?
abstract-algebra general-topology geometry algebraic-topology product-space
asked Nov 21 '18 at 14:32
user454042
235
|
show 3 more comments
In Hatcher prop 4.2 he proves that the n-th homotopy group of a product $Xtimes Y$ (for $X$ and $Y$ path-connected) is isomorphic to the product of the n-th homotopy groups of $X$ and $Y$. I wonder if a similar statement is true for relative homotopy groups? I.e. do we have a statement like
$pi_n(Xtimes Y, Atimes B, (x_0,y_0))=pi_n(X,A,x_0)times pi_n(Y,B,y_0)$?
abstract-algebra general-topology geometry algebraic-topology product-space
asked Nov 21 '18 at 14:32
user454042
235
2
I don't believe it holds like you have stated. Note that the product of pairs looks like $(X,A)times (Y,B)=(Xtimes Y,Xtimes Bcup Atimes Y)$, and I believe that it does hold that $pi_n(X,A)times pi_n(Y,B)cong pi_n((X,A)times (Y,B))$.
– Tyrone
Nov 21 '18 at 18:01
1
@Tyrone Your final formula is certainly correct: as you say it is just the universal property. And here is an example that shows OP's formula isn't right: take $A = X$ and $B = varnothing$. (If this feels too trivial take $B = *$.)
– Mike Miller
Nov 21 '18 at 18:11
1
@MikeMiller $B = emptyset$ is impossible since $y_0 in B$.
– Paul Frost
Nov 21 '18 at 23:24
@PaulFrost Good point, silly mistake. In any case, $B = y_0$ still presents a counterexample in most cases, the easiest being $X = Y = S^1$.
– Mike Miller
Nov 21 '18 at 23:35
@MikeMiller Please check my answer, I hope I didn't make a mistake.
– Paul Frost
Nov 22 '18 at 0:08
|
show 3 more comments
In Hatcher prop 4.2 he proves that the n-th homotopy group of a product $Xtimes Y$ (for $X$ and $Y$ path-connected) is isomorphic to the product of the n-th homotopy groups of $X$ and $Y$. I wonder if a similar statement is true for relative homotopy groups? I.e. do we have a statement like
$pi_n(Xtimes Y, Atimes B, (x_0,y_0))=pi_n(X,A,x_0)times pi_n(Y,B,y_0)$?
abstract-algebra general-topology geometry algebraic-topology product-space
asked Nov 21 '18 at 14:32
user454042
235
In Hatcher prop 4.2 he proves that the n-th homotopy group of a product $Xtimes Y$ (for $X$ and $Y$ path-connected) is isomorphic to the product of the n-th homotopy groups of $X$ and $Y$. I wonder if a similar statement is true for relative homotopy groups? I.e. do we have a statement like
$pi_n(Xtimes Y, Atimes B, (x_0,y_0))=pi_n(X,A,x_0)times pi_n(Y,B,y_0)$?
abstract-algebra general-topology geometry algebraic-topology product-space
abstract-algebra general-topology geometry algebraic-topology product-space
asked Nov 21 '18 at 14:32
user454042
235
asked Nov 21 '18 at 14:32
user454042
235
edited Nov 21 '18 at 15:37
asked Nov 21 '18 at 14:32
user454042
235
asked Nov 21 '18 at 14:32
user454042
235
asked Nov 21 '18 at 14:32
user454042
235
235
2
I don't believe it holds like you have stated. Note that the product of pairs looks like $(X,A)times (Y,B)=(Xtimes Y,Xtimes Bcup Atimes Y)$, and I believe that it does hold that $pi_n(X,A)times pi_n(Y,B)cong pi_n((X,A)times (Y,B))$.
– Tyrone
Nov 21 '18 at 18:01
1
@Tyrone Your final formula is certainly correct: as you say it is just the universal property. And here is an example that shows OP's formula isn't right: take $A = X$ and $B = varnothing$. (If this feels too trivial take $B = *$.)
– Mike Miller
Nov 21 '18 at 18:11
1
@MikeMiller $B = emptyset$ is impossible since $y_0 in B$.
– Paul Frost
Nov 21 '18 at 23:24
@PaulFrost Good point, silly mistake. In any case, $B = y_0$ still presents a counterexample in most cases, the easiest being $X = Y = S^1$.
– Mike Miller
Nov 21 '18 at 23:35
@MikeMiller Please check my answer, I hope I didn't make a mistake.
– Paul Frost
Nov 22 '18 at 0:08
|
show 3 more comments
2
I don't believe it holds like you have stated. Note that the product of pairs looks like $(X,A)times (Y,B)=(Xtimes Y,Xtimes Bcup Atimes Y)$, and I believe that it does hold that $pi_n(X,A)times pi_n(Y,B)cong pi_n((X,A)times (Y,B))$.
– Tyrone
Nov 21 '18 at 18:01
1
@Tyrone Your final formula is certainly correct: as you say it is just the universal property. And here is an example that shows OP's formula isn't right: take $A = X$ and $B = varnothing$. (If this feels too trivial take $B = *$.)
– Mike Miller
Nov 21 '18 at 18:11
1
@MikeMiller $B = emptyset$ is impossible since $y_0 in B$.
– Paul Frost
Nov 21 '18 at 23:24
@PaulFrost Good point, silly mistake. In any case, $B = y_0$ still presents a counterexample in most cases, the easiest being $X = Y = S^1$.
– Mike Miller
Nov 21 '18 at 23:35
@MikeMiller Please check my answer, I hope I didn't make a mistake.
– Paul Frost
Nov 22 '18 at 0:08
2
2
I don't believe it holds like you have stated. Note that the product of pairs looks like $(X,A)times (Y,B)=(Xtimes Y,Xtimes Bcup Atimes Y)$, and I believe that it does hold that $pi_n(X,A)times pi_n(Y,B)cong pi_n((X,A)times (Y,B))$.
– Tyrone
Nov 21 '18 at 18:01
I don't believe it holds like you have stated. Note that the product of pairs looks like $(X,A)times (Y,B)=(Xtimes Y,Xtimes Bcup Atimes Y)$, and I believe that it does hold that $pi_n(X,A)times pi_n(Y,B)cong pi_n((X,A)times (Y,B))$.
– Tyrone
Nov 21 '18 at 18:01
1
1
@Tyrone Your final formula is certainly correct: as you say it is just the universal property. And here is an example that shows OP's formula isn't right: take $A = X$ and $B = varnothing$. (If this feels too trivial take $B = *$.)
– Mike Miller
Nov 21 '18 at 18:11
@Tyrone Your final formula is certainly correct: as you say it is just the universal property. And here is an example that shows OP's formula isn't right: take $A = X$ and $B = varnothing$. (If this feels too trivial take $B = *$.)
– Mike Miller
Nov 21 '18 at 18:11
1
1
@MikeMiller $B = emptyset$ is impossible since $y_0 in B$.
– Paul Frost
Nov 21 '18 at 23:24
@MikeMiller $B = emptyset$ is impossible since $y_0 in B$.
– Paul Frost
Nov 21 '18 at 23:24
@PaulFrost Good point, silly mistake. In any case, $B = y_0$ still presents a counterexample in most cases, the easiest being $X = Y = S^1$.
– Mike Miller
Nov 21 '18 at 23:35
@PaulFrost Good point, silly mistake. In any case, $B = y_0$ still presents a counterexample in most cases, the easiest being $X = Y = S^1$.
– Mike Miller
Nov 21 '18 at 23:35
@MikeMiller Please check my answer, I hope I didn't make a mistake.
– Paul Frost
Nov 22 '18 at 0:08
@MikeMiller Please check my answer, I hope I didn't make a mistake.
– Paul Frost
Nov 22 '18 at 0:08
|
show 3 more comments
1 Answer
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$pi_n(Y,B,b_0)$ is the set of homotopy classes of maps $(D^n,S^{n-1},*) to (Y,B,b_0)$. Hatcher's proof of 4.2 applies verbatim to relative homotopy groups: Maps $(D^n,S^{n-1},*) to (X_ 1times X_2, A_1 times A_2, (a_1,a_2))$ can be identified with pairs of maps $(D^n,S^{n-1},*) to (X_ i, A_i,a_i)$, and the same is true for homotopies.
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
So you are saying the formula I stated is true? I‘m confused because of the counterexample of the comments.
– user454042
Nov 22 '18 at 9:11
Yes, it is true. The comments were based on a definition of $(X,A) times (Y,B)$ which does not agree with that in your question.
– Paul Frost
Nov 22 '18 at 21:15
@usee454042 I was just wrong.
– Mike Miller
Nov 23 '18 at 1:27
add a comment |
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$pi_n(Y,B,b_0)$ is the set of homotopy classes of maps $(D^n,S^{n-1},*) to (Y,B,b_0)$. Hatcher's proof of 4.2 applies verbatim to relative homotopy groups: Maps $(D^n,S^{n-1},*) to (X_ 1times X_2, A_1 times A_2, (a_1,a_2))$ can be identified with pairs of maps $(D^n,S^{n-1},*) to (X_ i, A_i,a_i)$, and the same is true for homotopies.
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
So you are saying the formula I stated is true? I‘m confused because of the counterexample of the comments.
– user454042
Nov 22 '18 at 9:11
Yes, it is true. The comments were based on a definition of $(X,A) times (Y,B)$ which does not agree with that in your question.
– Paul Frost
Nov 22 '18 at 21:15
@usee454042 I was just wrong.
– Mike Miller
Nov 23 '18 at 1:27
add a comment |
$pi_n(Y,B,b_0)$ is the set of homotopy classes of maps $(D^n,S^{n-1},*) to (Y,B,b_0)$. Hatcher's proof of 4.2 applies verbatim to relative homotopy groups: Maps $(D^n,S^{n-1},*) to (X_ 1times X_2, A_1 times A_2, (a_1,a_2))$ can be identified with pairs of maps $(D^n,S^{n-1},*) to (X_ i, A_i,a_i)$, and the same is true for homotopies.
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
So you are saying the formula I stated is true? I‘m confused because of the counterexample of the comments.
– user454042
Nov 22 '18 at 9:11
Yes, it is true. The comments were based on a definition of $(X,A) times (Y,B)$ which does not agree with that in your question.
– Paul Frost
Nov 22 '18 at 21:15
@usee454042 I was just wrong.
– Mike Miller
Nov 23 '18 at 1:27
add a comment |
$pi_n(Y,B,b_0)$ is the set of homotopy classes of maps $(D^n,S^{n-1},*) to (Y,B,b_0)$. Hatcher's proof of 4.2 applies verbatim to relative homotopy groups: Maps $(D^n,S^{n-1},*) to (X_ 1times X_2, A_1 times A_2, (a_1,a_2))$ can be identified with pairs of maps $(D^n,S^{n-1},*) to (X_ i, A_i,a_i)$, and the same is true for homotopies.
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
$pi_n(Y,B,b_0)$ is the set of homotopy classes of maps $(D^n,S^{n-1},*) to (Y,B,b_0)$. Hatcher's proof of 4.2 applies verbatim to relative homotopy groups: Maps $(D^n,S^{n-1},*) to (X_ 1times X_2, A_1 times A_2, (a_1,a_2))$ can be identified with pairs of maps $(D^n,S^{n-1},*) to (X_ i, A_i,a_i)$, and the same is true for homotopies.
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
edited Nov 22 '18 at 22:13
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
answered Nov 22 '18 at 0:31
Paul Frost
9,3762631
9,3762631
So you are saying the formula I stated is true? I‘m confused because of the counterexample of the comments.
– user454042
Nov 22 '18 at 9:11
Yes, it is true. The comments were based on a definition of $(X,A) times (Y,B)$ which does not agree with that in your question.
– Paul Frost
Nov 22 '18 at 21:15
@usee454042 I was just wrong.
– Mike Miller
Nov 23 '18 at 1:27
add a comment |
So you are saying the formula I stated is true? I‘m confused because of the counterexample of the comments.
– user454042
Nov 22 '18 at 9:11
Yes, it is true. The comments were based on a definition of $(X,A) times (Y,B)$ which does not agree with that in your question.
– Paul Frost
Nov 22 '18 at 21:15
@usee454042 I was just wrong.
– Mike Miller
Nov 23 '18 at 1:27
So you are saying the formula I stated is true? I‘m confused because of the counterexample of the comments.
– user454042
Nov 22 '18 at 9:11
So you are saying the formula I stated is true? I‘m confused because of the counterexample of the comments.
– user454042
Nov 22 '18 at 9:11
Yes, it is true. The comments were based on a definition of $(X,A) times (Y,B)$ which does not agree with that in your question.
– Paul Frost
Nov 22 '18 at 21:15
Yes, it is true. The comments were based on a definition of $(X,A) times (Y,B)$ which does not agree with that in your question.
– Paul Frost
Nov 22 '18 at 21:15
@usee454042 I was just wrong.
– Mike Miller
Nov 23 '18 at 1:27
@usee454042 I was just wrong.
– Mike Miller
Nov 23 '18 at 1:27
add a comment |
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2
I don't believe it holds like you have stated. Note that the product of pairs looks like $(X,A)times (Y,B)=(Xtimes Y,Xtimes Bcup Atimes Y)$, and I believe that it does hold that $pi_n(X,A)times pi_n(Y,B)cong pi_n((X,A)times (Y,B))$.
– Tyrone
Nov 21 '18 at 18:01
1
@Tyrone Your final formula is certainly correct: as you say it is just the universal property. And here is an example that shows OP's formula isn't right: take $A = X$ and $B = varnothing$. (If this feels too trivial take $B = *$.)
– Mike Miller
Nov 21 '18 at 18:11
1
@MikeMiller $B = emptyset$ is impossible since $y_0 in B$.
– Paul Frost
Nov 21 '18 at 23:24
@PaulFrost Good point, silly mistake. In any case, $B = y_0$ still presents a counterexample in most cases, the easiest being $X = Y = S^1$.
– Mike Miller
Nov 21 '18 at 23:35
@MikeMiller Please check my answer, I hope I didn't make a mistake.
– Paul Frost
Nov 22 '18 at 0:08