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Left multiplication is homeomorphism of topological groups
The Next CEO of Stack OverflowGroup 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
32011
$endgroup$
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
32011
$endgroup$
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
32011
$endgroup$
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
32011
asked Mar 25 at 15:24
LBJFSLBJFS
32011
edited Mar 25 at 15:53
LBJFS
asked Mar 25 at 15:24
LBJFSLBJFS
32011
asked Mar 25 at 15:24
LBJFSLBJFS
32011
asked Mar 25 at 15:24
LBJFSLBJFS
32011
32011
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
4,5762727
$endgroup$
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
171k23132240
$endgroup$
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
7801314
$endgroup$
$begingroup$
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
24k23482
$endgroup$
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
4,5762727
$endgroup$
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
4,5762727
$endgroup$
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
4,5762727
$endgroup$
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
4,5762727
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,5762727
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,5762727
answered Mar 25 at 15:29
Thomas ShelbyThomas Shelby
4,5762727
4,5762727
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
171k23132240
$endgroup$
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
171k23132240
$endgroup$
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
171k23132240
$endgroup$
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
171k23132240
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 25 at 15:26
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
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
7801314
$endgroup$
$begingroup$
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
7801314
$endgroup$
$begingroup$
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
7801314
$endgroup$
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
7801314
edited Mar 25 at 23:11
answered Mar 25 at 16:01
user458276user458276
7801314
answered Mar 25 at 16:01
user458276user458276
7801314
answered Mar 25 at 16:01
user458276user458276
7801314
7801314
$begingroup$
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
24k23482
$endgroup$
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
24k23482
$endgroup$
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
24k23482
$endgroup$
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
24k23482
answered Mar 25 at 16:48
tomasztomasz
24k23482
answered Mar 25 at 16:48
tomasztomasz
24k23482
answered Mar 25 at 16:48
tomasztomasz
24k23482
24k23482
add a comment |
add a comment |
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$begingroup$
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
$begingroup$
In the end, it sufficies to notice that the restriction of a continuous function is continuous.
$endgroup$
– LBJFS
Mar 25 at 15:47
1
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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