Mixing
Does the center of $pi_1(Y)$ act trivial on $[X,Y]_star$?
$begingroup$
Let $X$ and $Y$ be based (and well-pointed) and connected. We have an action of $pi_1(Y)$ on the set $[X,Y]_star$ of based homotopy classes of based maps. The quotient is just the set $[X,Y]$ of free homotopy classes of free maps.
In the case $Y=BG$ with $G$ a discrete group, we know that $pi_1(Y)=G$ and $[X,Y]_star cong mathrm{Hom}(pi_1(X),G)$. The $G$-action on the left side (as described above) corresponds to conjugation on the right side. Therefore, the center of $pi_1(Y)$ acts trivially.
Is this true in general, even if $Y$ is not aspherical?
algebraic-topology homotopy-theory fundamental-groups classifying-spaces
asked Jan 21 at 9:29
FKranholdFKranhold
1987
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be based (and well-pointed) and connected. We have an action of $pi_1(Y)$ on the set $[X,Y]_star$ of based homotopy classes of based maps. The quotient is just the set $[X,Y]$ of free homotopy classes of free maps.
In the case $Y=BG$ with $G$ a discrete group, we know that $pi_1(Y)=G$ and $[X,Y]_star cong mathrm{Hom}(pi_1(X),G)$. The $G$-action on the left side (as described above) corresponds to conjugation on the right side. Therefore, the center of $pi_1(Y)$ acts trivially.
Is this true in general, even if $Y$ is not aspherical?
algebraic-topology homotopy-theory fundamental-groups classifying-spaces
asked Jan 21 at 9:29
FKranholdFKranhold
1987
$endgroup$
2
$begingroup$
Ah! No, sorry, my conjecture is wrong. Consider $Y=mathbb{S}^1veemathbb{S}^2$ and $X=mathbb{S}^2$. Then $pi_1(Y)$ is abelian, but acts nontrivially on $pi_2(Y)=[X,Y]_star$.
$endgroup$
– FKranhold
Jan 21 at 9:57
add a comment |
$begingroup$
Let $X$ and $Y$ be based (and well-pointed) and connected. We have an action of $pi_1(Y)$ on the set $[X,Y]_star$ of based homotopy classes of based maps. The quotient is just the set $[X,Y]$ of free homotopy classes of free maps.
In the case $Y=BG$ with $G$ a discrete group, we know that $pi_1(Y)=G$ and $[X,Y]_star cong mathrm{Hom}(pi_1(X),G)$. The $G$-action on the left side (as described above) corresponds to conjugation on the right side. Therefore, the center of $pi_1(Y)$ acts trivially.
Is this true in general, even if $Y$ is not aspherical?
algebraic-topology homotopy-theory fundamental-groups classifying-spaces
asked Jan 21 at 9:29
FKranholdFKranhold
1987
$endgroup$
Let $X$ and $Y$ be based (and well-pointed) and connected. We have an action of $pi_1(Y)$ on the set $[X,Y]_star$ of based homotopy classes of based maps. The quotient is just the set $[X,Y]$ of free homotopy classes of free maps.
In the case $Y=BG$ with $G$ a discrete group, we know that $pi_1(Y)=G$ and $[X,Y]_star cong mathrm{Hom}(pi_1(X),G)$. The $G$-action on the left side (as described above) corresponds to conjugation on the right side. Therefore, the center of $pi_1(Y)$ acts trivially.
Is this true in general, even if $Y$ is not aspherical?
algebraic-topology homotopy-theory fundamental-groups classifying-spaces
algebraic-topology homotopy-theory fundamental-groups classifying-spaces
asked Jan 21 at 9:29
FKranholdFKranhold
1987
asked Jan 21 at 9:29
FKranholdFKranhold
1987
asked Jan 21 at 9:29
FKranholdFKranhold
1987
asked Jan 21 at 9:29
FKranholdFKranhold
1987
asked Jan 21 at 9:29
FKranholdFKranhold
1987
1987
2
$begingroup$
Ah! No, sorry, my conjecture is wrong. Consider $Y=mathbb{S}^1veemathbb{S}^2$ and $X=mathbb{S}^2$. Then $pi_1(Y)$ is abelian, but acts nontrivially on $pi_2(Y)=[X,Y]_star$.
$endgroup$
– FKranhold
Jan 21 at 9:57
add a comment |
2
$begingroup$
Ah! No, sorry, my conjecture is wrong. Consider $Y=mathbb{S}^1veemathbb{S}^2$ and $X=mathbb{S}^2$. Then $pi_1(Y)$ is abelian, but acts nontrivially on $pi_2(Y)=[X,Y]_star$.
$endgroup$
– FKranhold
Jan 21 at 9:57
2
2
$begingroup$
Ah! No, sorry, my conjecture is wrong. Consider $Y=mathbb{S}^1veemathbb{S}^2$ and $X=mathbb{S}^2$. Then $pi_1(Y)$ is abelian, but acts nontrivially on $pi_2(Y)=[X,Y]_star$.
$endgroup$
– FKranhold
Jan 21 at 9:57
$begingroup$
Ah! No, sorry, my conjecture is wrong. Consider $Y=mathbb{S}^1veemathbb{S}^2$ and $X=mathbb{S}^2$. Then $pi_1(Y)$ is abelian, but acts nontrivially on $pi_2(Y)=[X,Y]_star$.
$endgroup$
– FKranhold
Jan 21 at 9:57
add a comment |
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$begingroup$
Ah! No, sorry, my conjecture is wrong. Consider $Y=mathbb{S}^1veemathbb{S}^2$ and $X=mathbb{S}^2$. Then $pi_1(Y)$ is abelian, but acts nontrivially on $pi_2(Y)=[X,Y]_star$.
$endgroup$
– FKranhold
Jan 21 at 9:57