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Riehl's Category Theory in Context - Exercise 1.3.ix.
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I'm working through the following exercise from Emily Riehl's Category Theory in Context,
Exercise 1.3.ix. For any group $G$, we may define other groups:
- the center $Z(G) = hg = gh forall g ∈ G$, a subgroup of $G$,
- the commutator subgroup $C(G)$, the subgroup of $G$ generated by elements $ghg^-1h^-1$ for any $g, h in G$, and
- the automorphism group $operatornameAut(G)$, the group of isomorphisms $phi: G rightarrow G$ in $operatornameGroup$.
Trivially, all three constructions define a functor from the discrete category of groups (with
only identity morphisms) to $operatornameGroup$. Are these constructions functorial in
the isomorphisms of groups? That is, do they extend to functors $operatornameGroup_iso rightarrow operatornameGroup$?
the epimorphisms of groups? That is, do they extend to functors $operatornameGroup_epi rightarrow operatornameGroup$?
all homomorphisms of groups? That is, do they extend to functors $operatornameGroup rightarrow operatornameGroup$?
I've concluded that for any group morphism $f : G rightarrow H$ in $operatornameGroup$, $f(Z(G)) subseteq Z(H)$ if $f$ is epi and $f(C(G))subseteq
C(H)$ for all $f$, and thus we have functors
$$
F : operatornameGroup_epi rightarrow operatornameGroup \
G mapsto Z(G) \
f mapsto frestriction_Z(G)^Z(H)
\
$$
$$
F' : operatornameGroup rightarrow operatornameGroup \
G mapsto C(G) \
f mapsto frestriction_C(G)^C(H)
$$
Now for the automorphisms, the trivial functor from the discrete category of groups to $operatornameGroup$ can be extended to $operatornameGroup_iso$ by taking the following functor:
$$
H : operatornameGroup_iso rightarrow operatornameGroup \
G mapsto operatornameAut(G) \
phi mapsto (f mapsto phi f phi^-1)
$$
so my question is:
Can we extend this construction any further?
It is not even clear to me if having a morphism $f : G rightarrow H$ guarantees the existence of a (non-trivial) morphism $operatornameAut(G) rightarrow operatornameAut(H)$ when $f$ is not an iso, and I'm thinking this is probably not the case. As pointed out in the comments, split epis seem to be a problem for an extension to $operatornameGroup$, although I haven't been able to find a counterexample yet and I am not sure that this will behave badly in $operatornameGroup_epi$, since split epis will be isomorphisms.
category-theory
asked yesterday
Guido A.
3,411523
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I'm working through the following exercise from Emily Riehl's Category Theory in Context,
Exercise 1.3.ix. For any group $G$, we may define other groups:
- the center $Z(G) = hg = gh forall g ∈ G$, a subgroup of $G$,
- the commutator subgroup $C(G)$, the subgroup of $G$ generated by elements $ghg^-1h^-1$ for any $g, h in G$, and
- the automorphism group $operatornameAut(G)$, the group of isomorphisms $phi: G rightarrow G$ in $operatornameGroup$.
Trivially, all three constructions define a functor from the discrete category of groups (with
only identity morphisms) to $operatornameGroup$. Are these constructions functorial in
the isomorphisms of groups? That is, do they extend to functors $operatornameGroup_iso rightarrow operatornameGroup$?
the epimorphisms of groups? That is, do they extend to functors $operatornameGroup_epi rightarrow operatornameGroup$?
all homomorphisms of groups? That is, do they extend to functors $operatornameGroup rightarrow operatornameGroup$?
I've concluded that for any group morphism $f : G rightarrow H$ in $operatornameGroup$, $f(Z(G)) subseteq Z(H)$ if $f$ is epi and $f(C(G))subseteq
C(H)$ for all $f$, and thus we have functors
$$
F : operatornameGroup_epi rightarrow operatornameGroup \
G mapsto Z(G) \
f mapsto frestriction_Z(G)^Z(H)
\
$$
$$
F' : operatornameGroup rightarrow operatornameGroup \
G mapsto C(G) \
f mapsto frestriction_C(G)^C(H)
$$
Now for the automorphisms, the trivial functor from the discrete category of groups to $operatornameGroup$ can be extended to $operatornameGroup_iso$ by taking the following functor:
$$
H : operatornameGroup_iso rightarrow operatornameGroup \
G mapsto operatornameAut(G) \
phi mapsto (f mapsto phi f phi^-1)
$$
so my question is:
Can we extend this construction any further?
It is not even clear to me if having a morphism $f : G rightarrow H$ guarantees the existence of a (non-trivial) morphism $operatornameAut(G) rightarrow operatornameAut(H)$ when $f$ is not an iso, and I'm thinking this is probably not the case. As pointed out in the comments, split epis seem to be a problem for an extension to $operatornameGroup$, although I haven't been able to find a counterexample yet and I am not sure that this will behave badly in $operatornameGroup_epi$, since split epis will be isomorphisms.
category-theory
asked yesterday
Guido A.
3,411523
Your conclusion is wrong for $Z(G)$; actually one can prove that there is no extension of $Z$ to $Group$. $Aut$ should have a similar issue
– Max
yesterday
I'm sorry, I had a lapsus, I meant to say we had a functor from $operatornameGroup_epi rightarrow operatornameGroup$ assigning $f$ to $f restriction_Z(G)^Z(H)$. Would you mind elaborating what is the issue that makes the construction impossible for automorphisms? Can we at least extend the functor to $operatornameGroup_epi$?
– Guido A.
yesterday
Ok, now that's correct. The issue with automorphisms is that there will often be split epimorphisms in $Group$ from groups with very different automorphism groups. I haven't thought about epis but I think there will also be issues there : essentially the same as for $Xmapsto mathfrakSX$
– Max
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
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I'm working through the following exercise from Emily Riehl's Category Theory in Context,
Exercise 1.3.ix. For any group $G$, we may define other groups:
- the center $Z(G) = hg = gh forall g ∈ G$, a subgroup of $G$,
- the commutator subgroup $C(G)$, the subgroup of $G$ generated by elements $ghg^-1h^-1$ for any $g, h in G$, and
- the automorphism group $operatornameAut(G)$, the group of isomorphisms $phi: G rightarrow G$ in $operatornameGroup$.
Trivially, all three constructions define a functor from the discrete category of groups (with
only identity morphisms) to $operatornameGroup$. Are these constructions functorial in
the isomorphisms of groups? That is, do they extend to functors $operatornameGroup_iso rightarrow operatornameGroup$?
the epimorphisms of groups? That is, do they extend to functors $operatornameGroup_epi rightarrow operatornameGroup$?
all homomorphisms of groups? That is, do they extend to functors $operatornameGroup rightarrow operatornameGroup$?
I've concluded that for any group morphism $f : G rightarrow H$ in $operatornameGroup$, $f(Z(G)) subseteq Z(H)$ if $f$ is epi and $f(C(G))subseteq
C(H)$ for all $f$, and thus we have functors
$$
F : operatornameGroup_epi rightarrow operatornameGroup \
G mapsto Z(G) \
f mapsto frestriction_Z(G)^Z(H)
\
$$
$$
F' : operatornameGroup rightarrow operatornameGroup \
G mapsto C(G) \
f mapsto frestriction_C(G)^C(H)
$$
Now for the automorphisms, the trivial functor from the discrete category of groups to $operatornameGroup$ can be extended to $operatornameGroup_iso$ by taking the following functor:
$$
H : operatornameGroup_iso rightarrow operatornameGroup \
G mapsto operatornameAut(G) \
phi mapsto (f mapsto phi f phi^-1)
$$
so my question is:
Can we extend this construction any further?
It is not even clear to me if having a morphism $f : G rightarrow H$ guarantees the existence of a (non-trivial) morphism $operatornameAut(G) rightarrow operatornameAut(H)$ when $f$ is not an iso, and I'm thinking this is probably not the case. As pointed out in the comments, split epis seem to be a problem for an extension to $operatornameGroup$, although I haven't been able to find a counterexample yet and I am not sure that this will behave badly in $operatornameGroup_epi$, since split epis will be isomorphisms.
category-theory
asked yesterday
Guido A.
3,411523
I'm working through the following exercise from Emily Riehl's Category Theory in Context,
Exercise 1.3.ix. For any group $G$, we may define other groups:
- the center $Z(G) = hg = gh forall g ∈ G$, a subgroup of $G$,
- the commutator subgroup $C(G)$, the subgroup of $G$ generated by elements $ghg^-1h^-1$ for any $g, h in G$, and
- the automorphism group $operatornameAut(G)$, the group of isomorphisms $phi: G rightarrow G$ in $operatornameGroup$.
Trivially, all three constructions define a functor from the discrete category of groups (with
only identity morphisms) to $operatornameGroup$. Are these constructions functorial in
the isomorphisms of groups? That is, do they extend to functors $operatornameGroup_iso rightarrow operatornameGroup$?
the epimorphisms of groups? That is, do they extend to functors $operatornameGroup_epi rightarrow operatornameGroup$?
all homomorphisms of groups? That is, do they extend to functors $operatornameGroup rightarrow operatornameGroup$?
I've concluded that for any group morphism $f : G rightarrow H$ in $operatornameGroup$, $f(Z(G)) subseteq Z(H)$ if $f$ is epi and $f(C(G))subseteq
C(H)$ for all $f$, and thus we have functors
$$
F : operatornameGroup_epi rightarrow operatornameGroup \
G mapsto Z(G) \
f mapsto frestriction_Z(G)^Z(H)
\
$$
$$
F' : operatornameGroup rightarrow operatornameGroup \
G mapsto C(G) \
f mapsto frestriction_C(G)^C(H)
$$
Now for the automorphisms, the trivial functor from the discrete category of groups to $operatornameGroup$ can be extended to $operatornameGroup_iso$ by taking the following functor:
$$
H : operatornameGroup_iso rightarrow operatornameGroup \
G mapsto operatornameAut(G) \
phi mapsto (f mapsto phi f phi^-1)
$$
so my question is:
Can we extend this construction any further?
It is not even clear to me if having a morphism $f : G rightarrow H$ guarantees the existence of a (non-trivial) morphism $operatornameAut(G) rightarrow operatornameAut(H)$ when $f$ is not an iso, and I'm thinking this is probably not the case. As pointed out in the comments, split epis seem to be a problem for an extension to $operatornameGroup$, although I haven't been able to find a counterexample yet and I am not sure that this will behave badly in $operatornameGroup_epi$, since split epis will be isomorphisms.
category-theory
asked yesterday
Guido A.
3,411523
edited 3 hours ago
asked yesterday
Guido A.
3,411523
asked yesterday
Guido A.
3,411523
asked yesterday
Guido A.
3,411523
3,411523
Your conclusion is wrong for $Z(G)$; actually one can prove that there is no extension of $Z$ to $Group$. $Aut$ should have a similar issue
– Max
yesterday
I'm sorry, I had a lapsus, I meant to say we had a functor from $operatornameGroup_epi rightarrow operatornameGroup$ assigning $f$ to $f restriction_Z(G)^Z(H)$. Would you mind elaborating what is the issue that makes the construction impossible for automorphisms? Can we at least extend the functor to $operatornameGroup_epi$?
– Guido A.
yesterday
Ok, now that's correct. The issue with automorphisms is that there will often be split epimorphisms in $Group$ from groups with very different automorphism groups. I haven't thought about epis but I think there will also be issues there : essentially the same as for $Xmapsto mathfrakSX$
– Max
yesterday
add a comment |Â
Your conclusion is wrong for $Z(G)$; actually one can prove that there is no extension of $Z$ to $Group$. $Aut$ should have a similar issue
– Max
yesterday
I'm sorry, I had a lapsus, I meant to say we had a functor from $operatornameGroup_epi rightarrow operatornameGroup$ assigning $f$ to $f restriction_Z(G)^Z(H)$. Would you mind elaborating what is the issue that makes the construction impossible for automorphisms? Can we at least extend the functor to $operatornameGroup_epi$?
– Guido A.
yesterday
Ok, now that's correct. The issue with automorphisms is that there will often be split epimorphisms in $Group$ from groups with very different automorphism groups. I haven't thought about epis but I think there will also be issues there : essentially the same as for $Xmapsto mathfrakSX$
– Max
yesterday
Your conclusion is wrong for $Z(G)$; actually one can prove that there is no extension of $Z$ to $Group$. $Aut$ should have a similar issue
– Max
yesterday
Your conclusion is wrong for $Z(G)$; actually one can prove that there is no extension of $Z$ to $Group$. $Aut$ should have a similar issue
– Max
yesterday
I'm sorry, I had a lapsus, I meant to say we had a functor from $operatornameGroup_epi rightarrow operatornameGroup$ assigning $f$ to $f restriction_Z(G)^Z(H)$. Would you mind elaborating what is the issue that makes the construction impossible for automorphisms? Can we at least extend the functor to $operatornameGroup_epi$?
– Guido A.
yesterday
I'm sorry, I had a lapsus, I meant to say we had a functor from $operatornameGroup_epi rightarrow operatornameGroup$ assigning $f$ to $f restriction_Z(G)^Z(H)$. Would you mind elaborating what is the issue that makes the construction impossible for automorphisms? Can we at least extend the functor to $operatornameGroup_epi$?
– Guido A.
yesterday
Ok, now that's correct. The issue with automorphisms is that there will often be split epimorphisms in $Group$ from groups with very different automorphism groups. I haven't thought about epis but I think there will also be issues there : essentially the same as for $Xmapsto mathfrakSX$
– Max
yesterday
Ok, now that's correct. The issue with automorphisms is that there will often be split epimorphisms in $Group$ from groups with very different automorphism groups. I haven't thought about epis but I think there will also be issues there : essentially the same as for $Xmapsto mathfrakSX$
– Max
yesterday
add a comment |Â
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Your conclusion is wrong for $Z(G)$; actually one can prove that there is no extension of $Z$ to $Group$. $Aut$ should have a similar issue
– Max
yesterday
I'm sorry, I had a lapsus, I meant to say we had a functor from $operatornameGroup_epi rightarrow operatornameGroup$ assigning $f$ to $f restriction_Z(G)^Z(H)$. Would you mind elaborating what is the issue that makes the construction impossible for automorphisms? Can we at least extend the functor to $operatornameGroup_epi$?
– Guido A.
yesterday
Ok, now that's correct. The issue with automorphisms is that there will often be split epimorphisms in $Group$ from groups with very different automorphism groups. I haven't thought about epis but I think there will also be issues there : essentially the same as for $Xmapsto mathfrakSX$
– Max
yesterday