Left multiplication is homeomorphism of topological groups Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraGroup topology and free topological groupsAxiomatizing topology through continuous mapsUniform continuity of scalar multiplication in topological vector spacesCusp is homeomorphic to $mathbbR^n=1$Are topological fiber bundles on the same base with homeomorphic fibers isomorphic?Topological group of units of a topological ringOne-point compactification of disjoint union is homeomorphic to wedge sum of one-point compactificationsQuotient map from topological group onto left cosets is openProve that the map $nmapsto 1/n, infty mapsto 0$ is a homeomorphismMaking the subgroup of units of a topological ring a topological group
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Left multiplication is homeomorphism of topological groups
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraGroup topology and free topological groupsAxiomatizing topology through continuous mapsUniform continuity of scalar multiplication in topological vector spacesCusp is homeomorphic to $mathbbR^n=1$Are topological fiber bundles on the same base with homeomorphic fibers isomorphic?Topological group of units of a topological ringOne-point compactification of disjoint union is homeomorphic to wedge sum of one-point compactificationsQuotient map from topological group onto left cosets is openProve that the map $nmapsto 1/n, infty mapsto 0$ is a homeomorphismMaking the subgroup of units of a topological ring a topological group
$begingroup$
This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.
To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.
Here I stuck. Can I conclude it is open? If yes, why?
Thank you in advance for your help.
general-topology continuity topological-groups
asked Mar 25 at 15:24
LBJFSLBJFS
416213
$endgroup$
add a comment |
$begingroup$
This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.
To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.
Here I stuck. Can I conclude it is open? If yes, why?
Thank you in advance for your help.
general-topology continuity topological-groups
asked Mar 25 at 15:24
LBJFSLBJFS
416213
$endgroup$
1
$begingroup$
Just notice left multiplication is a restiction of the multiplication function to the subspace $atimes G$
$endgroup$
– YuiTo Cheng
Mar 25 at 15:28
1
$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
Mar 25 at 15:47
1
$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
Mar 25 at 15:50
$begingroup$
Thank you very much!
$endgroup$
– LBJFS
Mar 25 at 15:52
add a comment |
$begingroup$
This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.
To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.
Here I stuck. Can I conclude it is open? If yes, why?
Thank you in advance for your help.
general-topology continuity topological-groups
asked Mar 25 at 15:24
LBJFSLBJFS
416213
$endgroup$
This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ is a homeomorphism of $G$.
Clearly, this map is bijective and it sufficies to show its continuity.
To prove continuity, if $U(ag)$ is an open nhbd of $agin G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)subseteq U(ag)$, from which I get $aW(g)subseteq U(ag)$.
Here I stuck. Can I conclude it is open? If yes, why?
Thank you in advance for your help.
general-topology continuity topological-groups
general-topology continuity topological-groups
asked Mar 25 at 15:24
LBJFSLBJFS
416213
asked Mar 25 at 15:24
LBJFSLBJFS
416213
edited Mar 25 at 15:53
LBJFS
asked Mar 25 at 15:24
LBJFSLBJFS
416213
asked Mar 25 at 15:24
LBJFSLBJFS
416213
asked Mar 25 at 15:24
LBJFSLBJFS
416213
416213
1
$begingroup$
Just notice left multiplication is a restiction of the multiplication function to the subspace $atimes G$
$endgroup$
– YuiTo Cheng
Mar 25 at 15:28
1
$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
Mar 25 at 15:47
1
$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
Mar 25 at 15:50
$begingroup$
Thank you very much!
$endgroup$
– LBJFS
Mar 25 at 15:52
add a comment |
1
$begingroup$
Just notice left multiplication is a restiction of the multiplication function to the subspace $atimes G$
$endgroup$
– YuiTo Cheng
Mar 25 at 15:28
1
$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
Mar 25 at 15:47
1
$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
Mar 25 at 15:50
$begingroup$
Thank you very much!
$endgroup$
– LBJFS
Mar 25 at 15:52
1
1
$begingroup$
Just notice left multiplication is a restiction of the multiplication function to the subspace $atimes G$
$endgroup$
– YuiTo Cheng
Mar 25 at 15:28
$begingroup$
Just notice left multiplication is a restiction of the multiplication function to the subspace $atimes G$
$endgroup$
– YuiTo Cheng
Mar 25 at 15:28
1
1
$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
Mar 25 at 15:47
$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
Mar 25 at 15:47
1
1
$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
Mar 25 at 15:50
$begingroup$
And this is straightforward
$endgroup$
– YuiTo Cheng
Mar 25 at 15:50
$begingroup$
Thank you very much!
$endgroup$
– LBJFS
Mar 25 at 15:52
$begingroup$
Thank you very much!
$endgroup$
– LBJFS
Mar 25 at 15:52
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to $a×G $. So what can you conclude?
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
add a comment |
$begingroup$
The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.
And the inverse of $f_a$ is $f_a^-1$, which is continuous by the same reason.
Therefore, $f_a$ is a homeomorphism.
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
$endgroup$
add a comment |
$begingroup$
This is essentially rephrasing the other answer, but in a Category Theoretic perspective:
Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^-1:Grightarrow G$ expressing multiplication, identity, and inversion.
Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$
Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.
Together, we have the desired result.
answered Mar 25 at 16:01
user458276user458276
7431315
$endgroup$
$begingroup$
I don't know category theory, but I love different perspectives! Thank you very much.
$endgroup$
– LBJFS
Mar 25 at 16:05
$begingroup$
I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
$endgroup$
– tomasz
Mar 25 at 16:54
$begingroup$
@tomasz I elaborated a bit. Hopefully this makes more sense.
$endgroup$
– user458276
Mar 25 at 23:11
$begingroup$
I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
$endgroup$
– tomasz
Mar 26 at 11:41
add a comment |
$begingroup$
Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).
Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to $a×G $. So what can you conclude?
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
add a comment |
$begingroup$
Hint:
For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to $a×G $. So what can you conclude?
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
add a comment |
$begingroup$
Hint:
For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to $a×G $. So what can you conclude?
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
$endgroup$
Hint:
For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to $a×G $. So what can you conclude?
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,8082727
4,8082727
add a comment |
add a comment |
$begingroup$
The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.
And the inverse of $f_a$ is $f_a^-1$, which is continuous by the same reason.
Therefore, $f_a$ is a homeomorphism.
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
$endgroup$
add a comment |
$begingroup$
The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.
And the inverse of $f_a$ is $f_a^-1$, which is continuous by the same reason.
Therefore, $f_a$ is a homeomorphism.
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
$endgroup$
add a comment |
$begingroup$
The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.
And the inverse of $f_a$ is $f_a^-1$, which is continuous by the same reason.
Therefore, $f_a$ is a homeomorphism.
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
$endgroup$
The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.
And the inverse of $f_a$ is $f_a^-1$, which is continuous by the same reason.
Therefore, $f_a$ is a homeomorphism.
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
177k24138250
177k24138250
add a comment |
add a comment |
$begingroup$
This is essentially rephrasing the other answer, but in a Category Theoretic perspective:
Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^-1:Grightarrow G$ expressing multiplication, identity, and inversion.
Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$
Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.
Together, we have the desired result.
answered Mar 25 at 16:01
user458276user458276
7431315
$endgroup$
$begingroup$
I don't know category theory, but I love different perspectives! Thank you very much.
$endgroup$
– LBJFS
Mar 25 at 16:05
$begingroup$
I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
$endgroup$
– tomasz
Mar 25 at 16:54
$begingroup$
@tomasz I elaborated a bit. Hopefully this makes more sense.
$endgroup$
– user458276
Mar 25 at 23:11
$begingroup$
I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
$endgroup$
– tomasz
Mar 26 at 11:41
add a comment |
$begingroup$
This is essentially rephrasing the other answer, but in a Category Theoretic perspective:
Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^-1:Grightarrow G$ expressing multiplication, identity, and inversion.
Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$
Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.
Together, we have the desired result.
answered Mar 25 at 16:01
user458276user458276
7431315
$endgroup$
$begingroup$
I don't know category theory, but I love different perspectives! Thank you very much.
$endgroup$
– LBJFS
Mar 25 at 16:05
$begingroup$
I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
$endgroup$
– tomasz
Mar 25 at 16:54
$begingroup$
@tomasz I elaborated a bit. Hopefully this makes more sense.
$endgroup$
– user458276
Mar 25 at 23:11
$begingroup$
I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
$endgroup$
– tomasz
Mar 26 at 11:41
add a comment |
$begingroup$
This is essentially rephrasing the other answer, but in a Category Theoretic perspective:
Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^-1:Grightarrow G$ expressing multiplication, identity, and inversion.
Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$
Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.
Together, we have the desired result.
answered Mar 25 at 16:01
user458276user458276
7431315
$endgroup$
This is essentially rephrasing the other answer, but in a Category Theoretic perspective:
Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^-1:Grightarrow G$ expressing multiplication, identity, and inversion.
Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ mapping $(g,h)mapsto g.$
Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.
Together, we have the desired result.
answered Mar 25 at 16:01
user458276user458276
7431315
edited Mar 25 at 23:11
answered Mar 25 at 16:01
user458276user458276
7431315
answered Mar 25 at 16:01
user458276user458276
7431315
answered Mar 25 at 16:01
user458276user458276
7431315
7431315
$begingroup$
I don't know category theory, but I love different perspectives! Thank you very much.
$endgroup$
– LBJFS
Mar 25 at 16:05
$begingroup$
I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
$endgroup$
– tomasz
Mar 25 at 16:54
$begingroup$
@tomasz I elaborated a bit. Hopefully this makes more sense.
$endgroup$
– user458276
Mar 25 at 23:11
$begingroup$
I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
$endgroup$
– tomasz
Mar 26 at 11:41
add a comment |
$begingroup$
I don't know category theory, but I love different perspectives! Thank you very much.
$endgroup$
– LBJFS
Mar 25 at 16:05
$begingroup$
I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
$endgroup$
– tomasz
Mar 25 at 16:54
$begingroup$
@tomasz I elaborated a bit. Hopefully this makes more sense.
$endgroup$
– user458276
Mar 25 at 23:11
$begingroup$
I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
$endgroup$
– tomasz
Mar 26 at 11:41
$begingroup$
I don't know category theory, but I love different perspectives! Thank you very much.
$endgroup$
– LBJFS
Mar 25 at 16:05
$begingroup$
I don't know category theory, but I love different perspectives! Thank you very much.
$endgroup$
– LBJFS
Mar 25 at 16:05
$begingroup$
I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
$endgroup$
– tomasz
Mar 25 at 16:54
$begingroup$
I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate?
$endgroup$
– tomasz
Mar 25 at 16:54
$begingroup$
@tomasz I elaborated a bit. Hopefully this makes more sense.
$endgroup$
– user458276
Mar 25 at 23:11
$begingroup$
@tomasz I elaborated a bit. Hopefully this makes more sense.
$endgroup$
– user458276
Mar 25 at 23:11
$begingroup$
I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
$endgroup$
– tomasz
Mar 26 at 11:41
$begingroup$
I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $Gto G^2$, $gmapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious.
$endgroup$
– tomasz
Mar 26 at 11:41
add a comment |
$begingroup$
Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).
Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
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Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).
Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
$endgroup$
add a comment |
$begingroup$
Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).
Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
$endgroup$
Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).
Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
answered Mar 25 at 16:48
tomasztomasz
24.1k23482
24.1k23482
add a comment |
add a comment |
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Just notice left multiplication is a restiction of the multiplication function to the subspace $atimes G$
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– YuiTo Cheng
Mar 25 at 15:28
1
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In the end, it sufficies to notice that the restriction of a continuous function is continuous.
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– LBJFS
Mar 25 at 15:47
1
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And this is straightforward
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– YuiTo Cheng
Mar 25 at 15:50
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Thank you very much!
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– LBJFS
Mar 25 at 15:52