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Do finite simplicial sets jointly detect isomorphisms in the homotopy category?
Clash Royale CLAN TAG#URR8PPP
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Let $mathcalH$ denote the homotopy category associated with the Kan-Quillen model structure on $mathbfsSet$. Suppose we have a map $fcolon X to Y$ between Kan complexes, such that for every finite simplicial set $K$ we have an isomorphism of the form: $$mathcalH(K,X) cong mathcalH(K,Y)$$ induced by postcomposition with $[f]$. Is it true that $f$ is then a weak equivalence?
homotopy-theory simplicial-stuff
asked 3 hours ago
Edoardo Lanari
564313
 |Â
show 8 more comments
up vote
5
down vote
favorite
Let $mathcalH$ denote the homotopy category associated with the Kan-Quillen model structure on $mathbfsSet$. Suppose we have a map $fcolon X to Y$ between Kan complexes, such that for every finite simplicial set $K$ we have an isomorphism of the form: $$mathcalH(K,X) cong mathcalH(K,Y)$$ induced by postcomposition with $[f]$. Is it true that $f$ is then a weak equivalence?
homotopy-theory simplicial-stuff
asked 3 hours ago
Edoardo Lanari
564313
Might be dumb, but there are finite models for all of the simplicial spheres, so shouldn't this imply that the map induces isomorphisms on all homotopy groups at all basepoints? Not sure if considering basepoints is kosher after going to the homotopy category.
– Harry Gindi
2 hours ago
Hi Harry. Yes, surely that is the intuition but these are unpointed mapping spaces, so how do you recover that? I was looking for a "conceptual" proof whatever that might mean, but I couldn't come up with a combinatorics one either.
– Edoardo Lanari
2 hours ago
2
I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version.
– Gasterbiter
2 hours ago
1
You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $pi_n$ by the action of $pi_1$. So it at least works in cases where the $pi_1$ action is trivial
– Simon Henry
1 hour ago
1
@SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups.
– Kevin Carlson
10 mins ago
 |Â
show 8 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $mathcalH$ denote the homotopy category associated with the Kan-Quillen model structure on $mathbfsSet$. Suppose we have a map $fcolon X to Y$ between Kan complexes, such that for every finite simplicial set $K$ we have an isomorphism of the form: $$mathcalH(K,X) cong mathcalH(K,Y)$$ induced by postcomposition with $[f]$. Is it true that $f$ is then a weak equivalence?
homotopy-theory simplicial-stuff
asked 3 hours ago
Edoardo Lanari
564313
Let $mathcalH$ denote the homotopy category associated with the Kan-Quillen model structure on $mathbfsSet$. Suppose we have a map $fcolon X to Y$ between Kan complexes, such that for every finite simplicial set $K$ we have an isomorphism of the form: $$mathcalH(K,X) cong mathcalH(K,Y)$$ induced by postcomposition with $[f]$. Is it true that $f$ is then a weak equivalence?
homotopy-theory simplicial-stuff
homotopy-theory simplicial-stuff
asked 3 hours ago
Edoardo Lanari
564313
asked 3 hours ago
Edoardo Lanari
564313
asked 3 hours ago
Edoardo Lanari
564313
asked 3 hours ago
Edoardo Lanari
564313
asked 3 hours ago
Edoardo Lanari
564313
564313
Might be dumb, but there are finite models for all of the simplicial spheres, so shouldn't this imply that the map induces isomorphisms on all homotopy groups at all basepoints? Not sure if considering basepoints is kosher after going to the homotopy category.
– Harry Gindi
2 hours ago
Hi Harry. Yes, surely that is the intuition but these are unpointed mapping spaces, so how do you recover that? I was looking for a "conceptual" proof whatever that might mean, but I couldn't come up with a combinatorics one either.
– Edoardo Lanari
2 hours ago
2
I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version.
– Gasterbiter
2 hours ago
1
You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $pi_n$ by the action of $pi_1$. So it at least works in cases where the $pi_1$ action is trivial
– Simon Henry
1 hour ago
1
@SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups.
– Kevin Carlson
10 mins ago
 |Â
show 8 more comments
Might be dumb, but there are finite models for all of the simplicial spheres, so shouldn't this imply that the map induces isomorphisms on all homotopy groups at all basepoints? Not sure if considering basepoints is kosher after going to the homotopy category.
– Harry Gindi
2 hours ago
Hi Harry. Yes, surely that is the intuition but these are unpointed mapping spaces, so how do you recover that? I was looking for a "conceptual" proof whatever that might mean, but I couldn't come up with a combinatorics one either.
– Edoardo Lanari
2 hours ago
2
I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version.
– Gasterbiter
2 hours ago
1
You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $pi_n$ by the action of $pi_1$. So it at least works in cases where the $pi_1$ action is trivial
– Simon Henry
1 hour ago
1
@SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups.
– Kevin Carlson
10 mins ago
Might be dumb, but there are finite models for all of the simplicial spheres, so shouldn't this imply that the map induces isomorphisms on all homotopy groups at all basepoints? Not sure if considering basepoints is kosher after going to the homotopy category.
– Harry Gindi
2 hours ago
Might be dumb, but there are finite models for all of the simplicial spheres, so shouldn't this imply that the map induces isomorphisms on all homotopy groups at all basepoints? Not sure if considering basepoints is kosher after going to the homotopy category.
– Harry Gindi
2 hours ago
Hi Harry. Yes, surely that is the intuition but these are unpointed mapping spaces, so how do you recover that? I was looking for a "conceptual" proof whatever that might mean, but I couldn't come up with a combinatorics one either.
– Edoardo Lanari
2 hours ago
Hi Harry. Yes, surely that is the intuition but these are unpointed mapping spaces, so how do you recover that? I was looking for a "conceptual" proof whatever that might mean, but I couldn't come up with a combinatorics one either.
– Edoardo Lanari
2 hours ago
2
2
I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version.
– Gasterbiter
2 hours ago
I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version.
– Gasterbiter
2 hours ago
1
1
You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $pi_n$ by the action of $pi_1$. So it at least works in cases where the $pi_1$ action is trivial
– Simon Henry
1 hour ago
You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $pi_n$ by the action of $pi_1$. So it at least works in cases where the $pi_1$ action is trivial
– Simon Henry
1 hour ago
1
1
@SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups.
– Kevin Carlson
10 mins ago
@SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups.
– Kevin Carlson
10 mins ago
 |Â
show 8 more comments
2 Answers
2
active
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up vote
2
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Nice question!
I would suspect that the existence of phantom maps (see the n-Lab entry on these for starters) would suggest that the answer is a lot more complex than one would suspect. I quote
''a continuous map $f$ from a CW-complex $X$ to a topological space $Y$ is a phantom map if it is not homotopic to a constant but for every finite CW-subcomplex $Zsubset X$ the restriction $f|_Z:Zto Y$ is homotopic to a constant."
The conditions you impose would imply that $X$ and $Y$ have the same $n$-type for all $n$. From an example of Adams in 1957, and in reply to an original question of J. H. C. Whitehead (both looking at CW-complexes rather than simplicial sets), we get that $X$ and $Y$ are not necessarily homotopy equivalent. (Do a search on `Spaces of the same n-type for all n' to get some papers relevant to this.) The obstruction is in a $lim^(1)$ group associated to the automorphisms of the spaces.
This is not quite an answer to that question since you are specifying that $f$ is given to start with, but is clearly related to your and Harry's doubts given that you are asking for unpointed mapping spaces.
A related problem occurs in Proper Homotopy Theory and there perhaps you can find a nice conceptual example as there are geometric interpretations of the $lim^(1)$ groups. The intuition is that the homotopies needed to give the isomorphisms $$mathcalH(K,X) cong mathcalH(K,Y)$$ might not be made homotopically coherent enough to build a homotopy inverse.
Clearly in your case you need $X$ and $Y$ to have non-trivial homotopy groups in all dimensions.
I hope this helps.
answered 1 hour ago
Tim Porter
7,19711732
Is it true if you ask instead for the induced map $X^Kto Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$?
– Harry Gindi
1 hour ago
4
Take $K=Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending.
– Tim Porter
1 hour ago
Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message.
– Edoardo Lanari
1 hour ago
@TimPorter Whoops.
– Harry Gindi
30 mins ago
It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional here-see the counterexample among classifying spaces of discrete groups linked in comments.
– Kevin Carlson
20 mins ago
 |Â
show 1 more comment
up vote
1
down vote
The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "half-exact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.
answered 12 mins ago
Karol Szumiło
6,0362030
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Nice question!
I would suspect that the existence of phantom maps (see the n-Lab entry on these for starters) would suggest that the answer is a lot more complex than one would suspect. I quote
''a continuous map $f$ from a CW-complex $X$ to a topological space $Y$ is a phantom map if it is not homotopic to a constant but for every finite CW-subcomplex $Zsubset X$ the restriction $f|_Z:Zto Y$ is homotopic to a constant."
The conditions you impose would imply that $X$ and $Y$ have the same $n$-type for all $n$. From an example of Adams in 1957, and in reply to an original question of J. H. C. Whitehead (both looking at CW-complexes rather than simplicial sets), we get that $X$ and $Y$ are not necessarily homotopy equivalent. (Do a search on `Spaces of the same n-type for all n' to get some papers relevant to this.) The obstruction is in a $lim^(1)$ group associated to the automorphisms of the spaces.
This is not quite an answer to that question since you are specifying that $f$ is given to start with, but is clearly related to your and Harry's doubts given that you are asking for unpointed mapping spaces.
A related problem occurs in Proper Homotopy Theory and there perhaps you can find a nice conceptual example as there are geometric interpretations of the $lim^(1)$ groups. The intuition is that the homotopies needed to give the isomorphisms $$mathcalH(K,X) cong mathcalH(K,Y)$$ might not be made homotopically coherent enough to build a homotopy inverse.
Clearly in your case you need $X$ and $Y$ to have non-trivial homotopy groups in all dimensions.
I hope this helps.
answered 1 hour ago
Tim Porter
7,19711732
Is it true if you ask instead for the induced map $X^Kto Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$?
– Harry Gindi
1 hour ago
4
Take $K=Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending.
– Tim Porter
1 hour ago
Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message.
– Edoardo Lanari
1 hour ago
@TimPorter Whoops.
– Harry Gindi
30 mins ago
It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional here-see the counterexample among classifying spaces of discrete groups linked in comments.
– Kevin Carlson
20 mins ago
 |Â
show 1 more comment
up vote
2
down vote
Nice question!
I would suspect that the existence of phantom maps (see the n-Lab entry on these for starters) would suggest that the answer is a lot more complex than one would suspect. I quote
''a continuous map $f$ from a CW-complex $X$ to a topological space $Y$ is a phantom map if it is not homotopic to a constant but for every finite CW-subcomplex $Zsubset X$ the restriction $f|_Z:Zto Y$ is homotopic to a constant."
The conditions you impose would imply that $X$ and $Y$ have the same $n$-type for all $n$. From an example of Adams in 1957, and in reply to an original question of J. H. C. Whitehead (both looking at CW-complexes rather than simplicial sets), we get that $X$ and $Y$ are not necessarily homotopy equivalent. (Do a search on `Spaces of the same n-type for all n' to get some papers relevant to this.) The obstruction is in a $lim^(1)$ group associated to the automorphisms of the spaces.
This is not quite an answer to that question since you are specifying that $f$ is given to start with, but is clearly related to your and Harry's doubts given that you are asking for unpointed mapping spaces.
A related problem occurs in Proper Homotopy Theory and there perhaps you can find a nice conceptual example as there are geometric interpretations of the $lim^(1)$ groups. The intuition is that the homotopies needed to give the isomorphisms $$mathcalH(K,X) cong mathcalH(K,Y)$$ might not be made homotopically coherent enough to build a homotopy inverse.
Clearly in your case you need $X$ and $Y$ to have non-trivial homotopy groups in all dimensions.
I hope this helps.
answered 1 hour ago
Tim Porter
7,19711732
Is it true if you ask instead for the induced map $X^Kto Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$?
– Harry Gindi
1 hour ago
4
Take $K=Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending.
– Tim Porter
1 hour ago
Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message.
– Edoardo Lanari
1 hour ago
@TimPorter Whoops.
– Harry Gindi
30 mins ago
It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional here-see the counterexample among classifying spaces of discrete groups linked in comments.
– Kevin Carlson
20 mins ago
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
Nice question!
I would suspect that the existence of phantom maps (see the n-Lab entry on these for starters) would suggest that the answer is a lot more complex than one would suspect. I quote
''a continuous map $f$ from a CW-complex $X$ to a topological space $Y$ is a phantom map if it is not homotopic to a constant but for every finite CW-subcomplex $Zsubset X$ the restriction $f|_Z:Zto Y$ is homotopic to a constant."
The conditions you impose would imply that $X$ and $Y$ have the same $n$-type for all $n$. From an example of Adams in 1957, and in reply to an original question of J. H. C. Whitehead (both looking at CW-complexes rather than simplicial sets), we get that $X$ and $Y$ are not necessarily homotopy equivalent. (Do a search on `Spaces of the same n-type for all n' to get some papers relevant to this.) The obstruction is in a $lim^(1)$ group associated to the automorphisms of the spaces.
This is not quite an answer to that question since you are specifying that $f$ is given to start with, but is clearly related to your and Harry's doubts given that you are asking for unpointed mapping spaces.
A related problem occurs in Proper Homotopy Theory and there perhaps you can find a nice conceptual example as there are geometric interpretations of the $lim^(1)$ groups. The intuition is that the homotopies needed to give the isomorphisms $$mathcalH(K,X) cong mathcalH(K,Y)$$ might not be made homotopically coherent enough to build a homotopy inverse.
Clearly in your case you need $X$ and $Y$ to have non-trivial homotopy groups in all dimensions.
I hope this helps.
answered 1 hour ago
Tim Porter
7,19711732
Nice question!
I would suspect that the existence of phantom maps (see the n-Lab entry on these for starters) would suggest that the answer is a lot more complex than one would suspect. I quote
''a continuous map $f$ from a CW-complex $X$ to a topological space $Y$ is a phantom map if it is not homotopic to a constant but for every finite CW-subcomplex $Zsubset X$ the restriction $f|_Z:Zto Y$ is homotopic to a constant."
The conditions you impose would imply that $X$ and $Y$ have the same $n$-type for all $n$. From an example of Adams in 1957, and in reply to an original question of J. H. C. Whitehead (both looking at CW-complexes rather than simplicial sets), we get that $X$ and $Y$ are not necessarily homotopy equivalent. (Do a search on `Spaces of the same n-type for all n' to get some papers relevant to this.) The obstruction is in a $lim^(1)$ group associated to the automorphisms of the spaces.
This is not quite an answer to that question since you are specifying that $f$ is given to start with, but is clearly related to your and Harry's doubts given that you are asking for unpointed mapping spaces.
A related problem occurs in Proper Homotopy Theory and there perhaps you can find a nice conceptual example as there are geometric interpretations of the $lim^(1)$ groups. The intuition is that the homotopies needed to give the isomorphisms $$mathcalH(K,X) cong mathcalH(K,Y)$$ might not be made homotopically coherent enough to build a homotopy inverse.
Clearly in your case you need $X$ and $Y$ to have non-trivial homotopy groups in all dimensions.
I hope this helps.
answered 1 hour ago
Tim Porter
7,19711732
edited 1 hour ago
answered 1 hour ago
Tim Porter
7,19711732
answered 1 hour ago
Tim Porter
7,19711732
answered 1 hour ago
Tim Porter
7,19711732
7,19711732
Is it true if you ask instead for the induced map $X^Kto Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$?
– Harry Gindi
1 hour ago
4
Take $K=Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending.
– Tim Porter
1 hour ago
Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message.
– Edoardo Lanari
1 hour ago
@TimPorter Whoops.
– Harry Gindi
30 mins ago
It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional here-see the counterexample among classifying spaces of discrete groups linked in comments.
– Kevin Carlson
20 mins ago
 |Â
show 1 more comment
Is it true if you ask instead for the induced map $X^Kto Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$?
– Harry Gindi
1 hour ago
4
Take $K=Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending.
– Tim Porter
1 hour ago
Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message.
– Edoardo Lanari
1 hour ago
@TimPorter Whoops.
– Harry Gindi
30 mins ago
It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional here-see the counterexample among classifying spaces of discrete groups linked in comments.
– Kevin Carlson
20 mins ago
Is it true if you ask instead for the induced map $X^Kto Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$?
– Harry Gindi
1 hour ago
Is it true if you ask instead for the induced map $X^Kto Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$?
– Harry Gindi
1 hour ago
4
4
Take $K=Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending.
– Tim Porter
1 hour ago
Take $K=Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending.
– Tim Porter
1 hour ago
Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message.
– Edoardo Lanari
1 hour ago
Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message.
– Edoardo Lanari
1 hour ago
@TimPorter Whoops.
– Harry Gindi
30 mins ago
@TimPorter Whoops.
– Harry Gindi
30 mins ago
It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional here-see the counterexample among classifying spaces of discrete groups linked in comments.
– Kevin Carlson
20 mins ago
It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional here-see the counterexample among classifying spaces of discrete groups linked in comments.
– Kevin Carlson
20 mins ago
 |Â
show 1 more comment
up vote
1
down vote
The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "half-exact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.
answered 12 mins ago
Karol Szumiło
6,0362030
add a comment |Â
up vote
1
down vote
The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "half-exact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.
answered 12 mins ago
Karol Szumiło
6,0362030
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "half-exact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.
answered 12 mins ago
Karol Szumiło
6,0362030
The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "half-exact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.
answered 12 mins ago
Karol Szumiło
6,0362030
answered 12 mins ago
Karol Szumiło
6,0362030
answered 12 mins ago
Karol Szumiło
6,0362030
answered 12 mins ago
Karol Szumiło
6,0362030
6,0362030
add a comment |Â
add a comment |Â
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Might be dumb, but there are finite models for all of the simplicial spheres, so shouldn't this imply that the map induces isomorphisms on all homotopy groups at all basepoints? Not sure if considering basepoints is kosher after going to the homotopy category.
– Harry Gindi
2 hours ago
Hi Harry. Yes, surely that is the intuition but these are unpointed mapping spaces, so how do you recover that? I was looking for a "conceptual" proof whatever that might mean, but I couldn't come up with a combinatorics one either.
– Edoardo Lanari
2 hours ago
2
I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version.
– Gasterbiter
2 hours ago
1
You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $pi_n$ by the action of $pi_1$. So it at least works in cases where the $pi_1$ action is trivial
– Simon Henry
1 hour ago
1
@SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups.
– Kevin Carlson
10 mins ago