Mixing
How to show that $text{Hom}_R(Atimes B ,M)cong text{Hom}_R(A,M)times text{Hom}_R(B,M) $ when $A, B$, and $M$...
I am working on the problem below.
Let $A,B$ and $M$ be $R-$mudules. Show that
(1) $text{Hom }_R(Atimes B,M)cong text{Hom }_R(A,M)times text{Hom }_R(B,M)$.
For $(1)$, I built a homomorphism $F:operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)rightarrow operatorname{Hom}_R(Atimes B,M)$ defined by $F(varphi_1,varphi_2)=varphi_1+varphi_2$.
It is well defined since $varphi_1+varphi_2=psi_1+psi_2$ whenever $(varphi_1,varphi_2)=(psi_1,psi_2)$.
Also, it is homomorphism since, $forall rin R$ $forall (varphi_1,varphi_2),(psi_1,psi_2)in operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)$,
begin{align*}
F((varphi_1(a),varphi_2(b))+r(psi_1(a),psi_2(b)))&=(varphi_1(a)+rpsi_1(a))+(varphi_2(b)+rpsi_2(b))\ &=(varphi_1(a)+varphi_2(b))+r(psi_1(a)+psi_2(b))\ &=F(varphi_1(a),varphi_2(b))+rF(psi_1(a),psi_2(b)).
end{align*}
$forall (a,b)in Atimes B$.
Let $Phiin operatorname{Hom}_R(Atimes B,M) $ be given and note that $Phi(cdot,0)in operatorname{Hom}_R(A,M)$ and $Phi(0,cdot)in operatorname{Hom}_R(B,M)$, and that for any $(a,b)in Atimes B$,
begin{align*}
F(Phi(a,0),Phi(0,b))=Phi(a,0)+Phi(0,b)=Phi(a,b).
end{align*}
Thus, $F$ is surjective.
Therefore, I only need to show that it is an injection. But I am having trouble in there. I just want to show that $ker(F)=0$ but it seems there are so many $varphiin operatorname{Hom}_R(A,M)$ and $psiin operatorname{Hom}_R(B,M)$ such that $varphi+psi=0$. Should I change the homomorphism I have built? It seems this $F$ is only reasonable one...
I thank for any help in advance.
abstract-algebra modules
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
add a comment |
I am working on the problem below.
Let $A,B$ and $M$ be $R-$mudules. Show that
(1) $text{Hom }_R(Atimes B,M)cong text{Hom }_R(A,M)times text{Hom }_R(B,M)$.
For $(1)$, I built a homomorphism $F:operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)rightarrow operatorname{Hom}_R(Atimes B,M)$ defined by $F(varphi_1,varphi_2)=varphi_1+varphi_2$.
It is well defined since $varphi_1+varphi_2=psi_1+psi_2$ whenever $(varphi_1,varphi_2)=(psi_1,psi_2)$.
Also, it is homomorphism since, $forall rin R$ $forall (varphi_1,varphi_2),(psi_1,psi_2)in operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)$,
begin{align*}
F((varphi_1(a),varphi_2(b))+r(psi_1(a),psi_2(b)))&=(varphi_1(a)+rpsi_1(a))+(varphi_2(b)+rpsi_2(b))\ &=(varphi_1(a)+varphi_2(b))+r(psi_1(a)+psi_2(b))\ &=F(varphi_1(a),varphi_2(b))+rF(psi_1(a),psi_2(b)).
end{align*}
$forall (a,b)in Atimes B$.
Let $Phiin operatorname{Hom}_R(Atimes B,M) $ be given and note that $Phi(cdot,0)in operatorname{Hom}_R(A,M)$ and $Phi(0,cdot)in operatorname{Hom}_R(B,M)$, and that for any $(a,b)in Atimes B$,
begin{align*}
F(Phi(a,0),Phi(0,b))=Phi(a,0)+Phi(0,b)=Phi(a,b).
end{align*}
Thus, $F$ is surjective.
Therefore, I only need to show that it is an injection. But I am having trouble in there. I just want to show that $ker(F)=0$ but it seems there are so many $varphiin operatorname{Hom}_R(A,M)$ and $psiin operatorname{Hom}_R(B,M)$ such that $varphi+psi=0$. Should I change the homomorphism I have built? It seems this $F$ is only reasonable one...
I thank for any help in advance.
abstract-algebra modules
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
1
You are making an assumption about what homomorphisms look like from $Atimes B$. Use your surjectivity argument to show that $Atimes B$ is the coproduct of $A$ and $B$. Then, given any pair of homomorphisms from $Ato M$ and $Bto M$ there will be a unique homomorphism from the coproduct to $M$.
– John Douma
Nov 21 '18 at 17:27
add a comment |
I am working on the problem below.
Let $A,B$ and $M$ be $R-$mudules. Show that
(1) $text{Hom }_R(Atimes B,M)cong text{Hom }_R(A,M)times text{Hom }_R(B,M)$.
For $(1)$, I built a homomorphism $F:operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)rightarrow operatorname{Hom}_R(Atimes B,M)$ defined by $F(varphi_1,varphi_2)=varphi_1+varphi_2$.
It is well defined since $varphi_1+varphi_2=psi_1+psi_2$ whenever $(varphi_1,varphi_2)=(psi_1,psi_2)$.
Also, it is homomorphism since, $forall rin R$ $forall (varphi_1,varphi_2),(psi_1,psi_2)in operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)$,
begin{align*}
F((varphi_1(a),varphi_2(b))+r(psi_1(a),psi_2(b)))&=(varphi_1(a)+rpsi_1(a))+(varphi_2(b)+rpsi_2(b))\ &=(varphi_1(a)+varphi_2(b))+r(psi_1(a)+psi_2(b))\ &=F(varphi_1(a),varphi_2(b))+rF(psi_1(a),psi_2(b)).
end{align*}
$forall (a,b)in Atimes B$.
Let $Phiin operatorname{Hom}_R(Atimes B,M) $ be given and note that $Phi(cdot,0)in operatorname{Hom}_R(A,M)$ and $Phi(0,cdot)in operatorname{Hom}_R(B,M)$, and that for any $(a,b)in Atimes B$,
begin{align*}
F(Phi(a,0),Phi(0,b))=Phi(a,0)+Phi(0,b)=Phi(a,b).
end{align*}
Thus, $F$ is surjective.
Therefore, I only need to show that it is an injection. But I am having trouble in there. I just want to show that $ker(F)=0$ but it seems there are so many $varphiin operatorname{Hom}_R(A,M)$ and $psiin operatorname{Hom}_R(B,M)$ such that $varphi+psi=0$. Should I change the homomorphism I have built? It seems this $F$ is only reasonable one...
I thank for any help in advance.
abstract-algebra modules
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
I am working on the problem below.
Let $A,B$ and $M$ be $R-$mudules. Show that
(1) $text{Hom }_R(Atimes B,M)cong text{Hom }_R(A,M)times text{Hom }_R(B,M)$.
For $(1)$, I built a homomorphism $F:operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)rightarrow operatorname{Hom}_R(Atimes B,M)$ defined by $F(varphi_1,varphi_2)=varphi_1+varphi_2$.
It is well defined since $varphi_1+varphi_2=psi_1+psi_2$ whenever $(varphi_1,varphi_2)=(psi_1,psi_2)$.
Also, it is homomorphism since, $forall rin R$ $forall (varphi_1,varphi_2),(psi_1,psi_2)in operatorname{Hom}_R(A,M)times operatorname{Hom}_R(B,M)$,
begin{align*}
F((varphi_1(a),varphi_2(b))+r(psi_1(a),psi_2(b)))&=(varphi_1(a)+rpsi_1(a))+(varphi_2(b)+rpsi_2(b))\ &=(varphi_1(a)+varphi_2(b))+r(psi_1(a)+psi_2(b))\ &=F(varphi_1(a),varphi_2(b))+rF(psi_1(a),psi_2(b)).
end{align*}
$forall (a,b)in Atimes B$.
Let $Phiin operatorname{Hom}_R(Atimes B,M) $ be given and note that $Phi(cdot,0)in operatorname{Hom}_R(A,M)$ and $Phi(0,cdot)in operatorname{Hom}_R(B,M)$, and that for any $(a,b)in Atimes B$,
begin{align*}
F(Phi(a,0),Phi(0,b))=Phi(a,0)+Phi(0,b)=Phi(a,b).
end{align*}
Thus, $F$ is surjective.
Therefore, I only need to show that it is an injection. But I am having trouble in there. I just want to show that $ker(F)=0$ but it seems there are so many $varphiin operatorname{Hom}_R(A,M)$ and $psiin operatorname{Hom}_R(B,M)$ such that $varphi+psi=0$. Should I change the homomorphism I have built? It seems this $F$ is only reasonable one...
I thank for any help in advance.
abstract-algebra modules
abstract-algebra modules
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
edited Nov 21 '18 at 19:44
Monstrous Moonshiner
2,25011337
2,25011337
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
asked Nov 21 '18 at 16:46
LeBLeB
1,061217
1,061217
1
You are making an assumption about what homomorphisms look like from $Atimes B$. Use your surjectivity argument to show that $Atimes B$ is the coproduct of $A$ and $B$. Then, given any pair of homomorphisms from $Ato M$ and $Bto M$ there will be a unique homomorphism from the coproduct to $M$.
– John Douma
Nov 21 '18 at 17:27
add a comment |
1
You are making an assumption about what homomorphisms look like from $Atimes B$. Use your surjectivity argument to show that $Atimes B$ is the coproduct of $A$ and $B$. Then, given any pair of homomorphisms from $Ato M$ and $Bto M$ there will be a unique homomorphism from the coproduct to $M$.
– John Douma
Nov 21 '18 at 17:27
1
1
You are making an assumption about what homomorphisms look like from $Atimes B$. Use your surjectivity argument to show that $Atimes B$ is the coproduct of $A$ and $B$. Then, given any pair of homomorphisms from $Ato M$ and $Bto M$ there will be a unique homomorphism from the coproduct to $M$.
– John Douma
Nov 21 '18 at 17:27
You are making an assumption about what homomorphisms look like from $Atimes B$. Use your surjectivity argument to show that $Atimes B$ is the coproduct of $A$ and $B$. Then, given any pair of homomorphisms from $Ato M$ and $Bto M$ there will be a unique homomorphism from the coproduct to $M$.
– John Douma
Nov 21 '18 at 17:27
add a comment |
3 Answers
3
active
oldest
votes
The map you are defining doesn't make sense. You don't need to worry about well-defined-ness of the map as there aren't any equivalence relations around to muck things up. Whatever you define that map to be at an element, will be fine because it not like there are a whole bunch of representatives of that element that could change the expression depending on which ones you pick.
More to the point however, it doesn't make sense to form the sum $varphi_1 + varphi_2$ when the domain of $varphi_1$ is $A$ and the domain of $varphi_2$ is $B$. Remember that addition of functions is typically defined pointwise, which can only make sense if those functions share the same domain. If you want to combine those two functions in a different manner than pointwise addition, then you need to indicate that by not using the addition symbol.
I'm not going to spend a lot of time reviewing that argument because it's a bit confusing and hard to read. But I can help guide you through the right process to show the necessary bijection. Given two functions $phi_1: A to M$ and $phi_2: B to M$, define $f(phi_1,phi_2)$ to be the function from $A times B$ to $M$ defined by $f(phi_1,phi_2)(a,b) = phi_1(a)+phi_2(b)$. Note that this is not a pointwise sum, and there is no need to argue about well-definedness.
Now we just need to exhibit an inverse map for $f$. In this case, this will be slicker than trying to argue for injectivity and surjectivity of $f$ directly. Given a function $phi: A times B to M$, we define $phi_1(a) = phi(a,0)$ and $phi_2(b) = phi(0,b)$ for all $a in A$, $b in B$. Then we define $g(phi)$ to be the ordered pair $(phi_1,phi_2) in operatorname{Hom}_R(A,M) times operatorname{Hom}_R(B,M)$. Then we have that $g(f(phi_1,phi_2) = g(phi) = (phi_1,phi_2)$ and $f(g(phi)) = f(phi_1,phi_2) = phi$ and so these maps really are inverses of each other. Therefore we are done.
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
add a comment |
Showing injectivity amounts to showing that the inverse map is well defined.
Let $F^{-1}$ be the inverse with $F(phi) = (phi_1, phi_2)$ with $phi_1(x)=phi(x, 0)$ and $phi_2(y) = phi(0,y)$. This map is obviousy well-defined, so $F$ is injective.
Edit: I realize that assuming $F^{-1}$ exists is begging the question. However, the proposition is easier to prove, in my opinion, starting with what I called $F^{-1}$.
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
add a comment |
Define the inclusion maps $i_A:Ato Atimes B$ by $i_A(a)=(a,0)$ and $i_B:Bto M$ by $i_B(b)=(0,b)$.
As you mentioned, these are module homomorphisms.
Given $fin Hom(Atimes B,M)$, Let $F(f)=(fcirc i_A, fcirc i_B)$. Since both components are the composition of homomorphisms, each component is a homomorphism.
Given $(phi, psi)in Hom(A,M)times Hom(B,M)$, define $G(phi,psi)$ by $G(phi,psi)(a,b)=phi(a)+psi(b)$.
Then $(Gcirc F)(f)=G(fcirc i_A, fcirc i_B)=fcirc i_A + fcirc i_B$.
$(fcirc i_A + fcirc i_B)(a,b)=fcirc i_A(a)+fcirc i_B(b)=f(a,0)+f(0,b)=f(a,b)$.
Therefore, $(Gcirc F)(f)=f$.
$(Fcirc G)(phi,psi)=F(phi+psi)=((phi+psi)i_A, ((phi+psi)i_B)$.
$(phi+psi)i_A(a)=(phi+psi)(a,0)=phi(a)$.
Likewise,$(phi+psi)i_B(b)=(phi+psi)(0,b)=psi(b)$.
Therefor, $(Fcirc G)(phi,psi)=(phi, psi)$.
Therefore, $F$ and $G$ are inverses of each other so they are isomorphisms.
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
add a comment |
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3 Answers
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3 Answers
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The map you are defining doesn't make sense. You don't need to worry about well-defined-ness of the map as there aren't any equivalence relations around to muck things up. Whatever you define that map to be at an element, will be fine because it not like there are a whole bunch of representatives of that element that could change the expression depending on which ones you pick.
More to the point however, it doesn't make sense to form the sum $varphi_1 + varphi_2$ when the domain of $varphi_1$ is $A$ and the domain of $varphi_2$ is $B$. Remember that addition of functions is typically defined pointwise, which can only make sense if those functions share the same domain. If you want to combine those two functions in a different manner than pointwise addition, then you need to indicate that by not using the addition symbol.
I'm not going to spend a lot of time reviewing that argument because it's a bit confusing and hard to read. But I can help guide you through the right process to show the necessary bijection. Given two functions $phi_1: A to M$ and $phi_2: B to M$, define $f(phi_1,phi_2)$ to be the function from $A times B$ to $M$ defined by $f(phi_1,phi_2)(a,b) = phi_1(a)+phi_2(b)$. Note that this is not a pointwise sum, and there is no need to argue about well-definedness.
Now we just need to exhibit an inverse map for $f$. In this case, this will be slicker than trying to argue for injectivity and surjectivity of $f$ directly. Given a function $phi: A times B to M$, we define $phi_1(a) = phi(a,0)$ and $phi_2(b) = phi(0,b)$ for all $a in A$, $b in B$. Then we define $g(phi)$ to be the ordered pair $(phi_1,phi_2) in operatorname{Hom}_R(A,M) times operatorname{Hom}_R(B,M)$. Then we have that $g(f(phi_1,phi_2) = g(phi) = (phi_1,phi_2)$ and $f(g(phi)) = f(phi_1,phi_2) = phi$ and so these maps really are inverses of each other. Therefore we are done.
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
add a comment |
The map you are defining doesn't make sense. You don't need to worry about well-defined-ness of the map as there aren't any equivalence relations around to muck things up. Whatever you define that map to be at an element, will be fine because it not like there are a whole bunch of representatives of that element that could change the expression depending on which ones you pick.
More to the point however, it doesn't make sense to form the sum $varphi_1 + varphi_2$ when the domain of $varphi_1$ is $A$ and the domain of $varphi_2$ is $B$. Remember that addition of functions is typically defined pointwise, which can only make sense if those functions share the same domain. If you want to combine those two functions in a different manner than pointwise addition, then you need to indicate that by not using the addition symbol.
I'm not going to spend a lot of time reviewing that argument because it's a bit confusing and hard to read. But I can help guide you through the right process to show the necessary bijection. Given two functions $phi_1: A to M$ and $phi_2: B to M$, define $f(phi_1,phi_2)$ to be the function from $A times B$ to $M$ defined by $f(phi_1,phi_2)(a,b) = phi_1(a)+phi_2(b)$. Note that this is not a pointwise sum, and there is no need to argue about well-definedness.
Now we just need to exhibit an inverse map for $f$. In this case, this will be slicker than trying to argue for injectivity and surjectivity of $f$ directly. Given a function $phi: A times B to M$, we define $phi_1(a) = phi(a,0)$ and $phi_2(b) = phi(0,b)$ for all $a in A$, $b in B$. Then we define $g(phi)$ to be the ordered pair $(phi_1,phi_2) in operatorname{Hom}_R(A,M) times operatorname{Hom}_R(B,M)$. Then we have that $g(f(phi_1,phi_2) = g(phi) = (phi_1,phi_2)$ and $f(g(phi)) = f(phi_1,phi_2) = phi$ and so these maps really are inverses of each other. Therefore we are done.
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
add a comment |
The map you are defining doesn't make sense. You don't need to worry about well-defined-ness of the map as there aren't any equivalence relations around to muck things up. Whatever you define that map to be at an element, will be fine because it not like there are a whole bunch of representatives of that element that could change the expression depending on which ones you pick.
More to the point however, it doesn't make sense to form the sum $varphi_1 + varphi_2$ when the domain of $varphi_1$ is $A$ and the domain of $varphi_2$ is $B$. Remember that addition of functions is typically defined pointwise, which can only make sense if those functions share the same domain. If you want to combine those two functions in a different manner than pointwise addition, then you need to indicate that by not using the addition symbol.
I'm not going to spend a lot of time reviewing that argument because it's a bit confusing and hard to read. But I can help guide you through the right process to show the necessary bijection. Given two functions $phi_1: A to M$ and $phi_2: B to M$, define $f(phi_1,phi_2)$ to be the function from $A times B$ to $M$ defined by $f(phi_1,phi_2)(a,b) = phi_1(a)+phi_2(b)$. Note that this is not a pointwise sum, and there is no need to argue about well-definedness.
Now we just need to exhibit an inverse map for $f$. In this case, this will be slicker than trying to argue for injectivity and surjectivity of $f$ directly. Given a function $phi: A times B to M$, we define $phi_1(a) = phi(a,0)$ and $phi_2(b) = phi(0,b)$ for all $a in A$, $b in B$. Then we define $g(phi)$ to be the ordered pair $(phi_1,phi_2) in operatorname{Hom}_R(A,M) times operatorname{Hom}_R(B,M)$. Then we have that $g(f(phi_1,phi_2) = g(phi) = (phi_1,phi_2)$ and $f(g(phi)) = f(phi_1,phi_2) = phi$ and so these maps really are inverses of each other. Therefore we are done.
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
The map you are defining doesn't make sense. You don't need to worry about well-defined-ness of the map as there aren't any equivalence relations around to muck things up. Whatever you define that map to be at an element, will be fine because it not like there are a whole bunch of representatives of that element that could change the expression depending on which ones you pick.
More to the point however, it doesn't make sense to form the sum $varphi_1 + varphi_2$ when the domain of $varphi_1$ is $A$ and the domain of $varphi_2$ is $B$. Remember that addition of functions is typically defined pointwise, which can only make sense if those functions share the same domain. If you want to combine those two functions in a different manner than pointwise addition, then you need to indicate that by not using the addition symbol.
I'm not going to spend a lot of time reviewing that argument because it's a bit confusing and hard to read. But I can help guide you through the right process to show the necessary bijection. Given two functions $phi_1: A to M$ and $phi_2: B to M$, define $f(phi_1,phi_2)$ to be the function from $A times B$ to $M$ defined by $f(phi_1,phi_2)(a,b) = phi_1(a)+phi_2(b)$. Note that this is not a pointwise sum, and there is no need to argue about well-definedness.
Now we just need to exhibit an inverse map for $f$. In this case, this will be slicker than trying to argue for injectivity and surjectivity of $f$ directly. Given a function $phi: A times B to M$, we define $phi_1(a) = phi(a,0)$ and $phi_2(b) = phi(0,b)$ for all $a in A$, $b in B$. Then we define $g(phi)$ to be the ordered pair $(phi_1,phi_2) in operatorname{Hom}_R(A,M) times operatorname{Hom}_R(B,M)$. Then we have that $g(f(phi_1,phi_2) = g(phi) = (phi_1,phi_2)$ and $f(g(phi)) = f(phi_1,phi_2) = phi$ and so these maps really are inverses of each other. Therefore we are done.
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
edited Nov 21 '18 at 20:48
quid♦
36.9k95093
36.9k95093
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
answered Nov 21 '18 at 19:38
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
2,25011337
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Showing injectivity amounts to showing that the inverse map is well defined.
Let $F^{-1}$ be the inverse with $F(phi) = (phi_1, phi_2)$ with $phi_1(x)=phi(x, 0)$ and $phi_2(y) = phi(0,y)$. This map is obviousy well-defined, so $F$ is injective.
Edit: I realize that assuming $F^{-1}$ exists is begging the question. However, the proposition is easier to prove, in my opinion, starting with what I called $F^{-1}$.
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
add a comment |
Showing injectivity amounts to showing that the inverse map is well defined.
Let $F^{-1}$ be the inverse with $F(phi) = (phi_1, phi_2)$ with $phi_1(x)=phi(x, 0)$ and $phi_2(y) = phi(0,y)$. This map is obviousy well-defined, so $F$ is injective.
Edit: I realize that assuming $F^{-1}$ exists is begging the question. However, the proposition is easier to prove, in my opinion, starting with what I called $F^{-1}$.
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
add a comment |
Showing injectivity amounts to showing that the inverse map is well defined.
Let $F^{-1}$ be the inverse with $F(phi) = (phi_1, phi_2)$ with $phi_1(x)=phi(x, 0)$ and $phi_2(y) = phi(0,y)$. This map is obviousy well-defined, so $F$ is injective.
Edit: I realize that assuming $F^{-1}$ exists is begging the question. However, the proposition is easier to prove, in my opinion, starting with what I called $F^{-1}$.
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
Showing injectivity amounts to showing that the inverse map is well defined.
Let $F^{-1}$ be the inverse with $F(phi) = (phi_1, phi_2)$ with $phi_1(x)=phi(x, 0)$ and $phi_2(y) = phi(0,y)$. This map is obviousy well-defined, so $F$ is injective.
Edit: I realize that assuming $F^{-1}$ exists is begging the question. However, the proposition is easier to prove, in my opinion, starting with what I called $F^{-1}$.
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
edited Nov 21 '18 at 19:46
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
answered Nov 21 '18 at 19:28
Lukas KoflerLukas Kofler
1
1
add a comment |
add a comment |
Define the inclusion maps $i_A:Ato Atimes B$ by $i_A(a)=(a,0)$ and $i_B:Bto M$ by $i_B(b)=(0,b)$.
As you mentioned, these are module homomorphisms.
Given $fin Hom(Atimes B,M)$, Let $F(f)=(fcirc i_A, fcirc i_B)$. Since both components are the composition of homomorphisms, each component is a homomorphism.
Given $(phi, psi)in Hom(A,M)times Hom(B,M)$, define $G(phi,psi)$ by $G(phi,psi)(a,b)=phi(a)+psi(b)$.
Then $(Gcirc F)(f)=G(fcirc i_A, fcirc i_B)=fcirc i_A + fcirc i_B$.
$(fcirc i_A + fcirc i_B)(a,b)=fcirc i_A(a)+fcirc i_B(b)=f(a,0)+f(0,b)=f(a,b)$.
Therefore, $(Gcirc F)(f)=f$.
$(Fcirc G)(phi,psi)=F(phi+psi)=((phi+psi)i_A, ((phi+psi)i_B)$.
$(phi+psi)i_A(a)=(phi+psi)(a,0)=phi(a)$.
Likewise,$(phi+psi)i_B(b)=(phi+psi)(0,b)=psi(b)$.
Therefor, $(Fcirc G)(phi,psi)=(phi, psi)$.
Therefore, $F$ and $G$ are inverses of each other so they are isomorphisms.
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
add a comment |
Define the inclusion maps $i_A:Ato Atimes B$ by $i_A(a)=(a,0)$ and $i_B:Bto M$ by $i_B(b)=(0,b)$.
As you mentioned, these are module homomorphisms.
Given $fin Hom(Atimes B,M)$, Let $F(f)=(fcirc i_A, fcirc i_B)$. Since both components are the composition of homomorphisms, each component is a homomorphism.
Given $(phi, psi)in Hom(A,M)times Hom(B,M)$, define $G(phi,psi)$ by $G(phi,psi)(a,b)=phi(a)+psi(b)$.
Then $(Gcirc F)(f)=G(fcirc i_A, fcirc i_B)=fcirc i_A + fcirc i_B$.
$(fcirc i_A + fcirc i_B)(a,b)=fcirc i_A(a)+fcirc i_B(b)=f(a,0)+f(0,b)=f(a,b)$.
Therefore, $(Gcirc F)(f)=f$.
$(Fcirc G)(phi,psi)=F(phi+psi)=((phi+psi)i_A, ((phi+psi)i_B)$.
$(phi+psi)i_A(a)=(phi+psi)(a,0)=phi(a)$.
Likewise,$(phi+psi)i_B(b)=(phi+psi)(0,b)=psi(b)$.
Therefor, $(Fcirc G)(phi,psi)=(phi, psi)$.
Therefore, $F$ and $G$ are inverses of each other so they are isomorphisms.
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
add a comment |
Define the inclusion maps $i_A:Ato Atimes B$ by $i_A(a)=(a,0)$ and $i_B:Bto M$ by $i_B(b)=(0,b)$.
As you mentioned, these are module homomorphisms.
Given $fin Hom(Atimes B,M)$, Let $F(f)=(fcirc i_A, fcirc i_B)$. Since both components are the composition of homomorphisms, each component is a homomorphism.
Given $(phi, psi)in Hom(A,M)times Hom(B,M)$, define $G(phi,psi)$ by $G(phi,psi)(a,b)=phi(a)+psi(b)$.
Then $(Gcirc F)(f)=G(fcirc i_A, fcirc i_B)=fcirc i_A + fcirc i_B$.
$(fcirc i_A + fcirc i_B)(a,b)=fcirc i_A(a)+fcirc i_B(b)=f(a,0)+f(0,b)=f(a,b)$.
Therefore, $(Gcirc F)(f)=f$.
$(Fcirc G)(phi,psi)=F(phi+psi)=((phi+psi)i_A, ((phi+psi)i_B)$.
$(phi+psi)i_A(a)=(phi+psi)(a,0)=phi(a)$.
Likewise,$(phi+psi)i_B(b)=(phi+psi)(0,b)=psi(b)$.
Therefor, $(Fcirc G)(phi,psi)=(phi, psi)$.
Therefore, $F$ and $G$ are inverses of each other so they are isomorphisms.
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
Define the inclusion maps $i_A:Ato Atimes B$ by $i_A(a)=(a,0)$ and $i_B:Bto M$ by $i_B(b)=(0,b)$.
As you mentioned, these are module homomorphisms.
Given $fin Hom(Atimes B,M)$, Let $F(f)=(fcirc i_A, fcirc i_B)$. Since both components are the composition of homomorphisms, each component is a homomorphism.
Given $(phi, psi)in Hom(A,M)times Hom(B,M)$, define $G(phi,psi)$ by $G(phi,psi)(a,b)=phi(a)+psi(b)$.
Then $(Gcirc F)(f)=G(fcirc i_A, fcirc i_B)=fcirc i_A + fcirc i_B$.
$(fcirc i_A + fcirc i_B)(a,b)=fcirc i_A(a)+fcirc i_B(b)=f(a,0)+f(0,b)=f(a,b)$.
Therefore, $(Gcirc F)(f)=f$.
$(Fcirc G)(phi,psi)=F(phi+psi)=((phi+psi)i_A, ((phi+psi)i_B)$.
$(phi+psi)i_A(a)=(phi+psi)(a,0)=phi(a)$.
Likewise,$(phi+psi)i_B(b)=(phi+psi)(0,b)=psi(b)$.
Therefor, $(Fcirc G)(phi,psi)=(phi, psi)$.
Therefore, $F$ and $G$ are inverses of each other so they are isomorphisms.
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
edited Nov 21 '18 at 20:45
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
answered Nov 21 '18 at 18:41
John DoumaJohn Douma
5,40211319
5,40211319
add a comment |
add a comment |
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You are making an assumption about what homomorphisms look like from $Atimes B$. Use your surjectivity argument to show that $Atimes B$ is the coproduct of $A$ and $B$. Then, given any pair of homomorphisms from $Ato M$ and $Bto M$ there will be a unique homomorphism from the coproduct to $M$.
– John Douma
Nov 21 '18 at 17:27