Mixing
$Hom_k(V,V otimes_k W) cong Hom_k(V,V)otimes_k W$
$begingroup$
I am thinking the following statement is true:
Let $V,W$ be finite dimensional $k$-vector spaces. Then we have
$$ Hom_k(V,V otimes_kW)cong Hom_k(V,V) otimes_kW$$
My attempt: EDIT ( I noticed I have typed up some nonsense)
There is a map from RHS to LHS, sending $f otimes_k w$ to element
$$ (f otimes_k w) (v) = f(v) otimes _k w $$
Motivation: this is an attempt to solve the more general question
abstract-algebra tensor-products
asked Jan 13 at 16:42
CL.CL.
2,2432825
$endgroup$
add a comment |
$begingroup$
I am thinking the following statement is true:
Let $V,W$ be finite dimensional $k$-vector spaces. Then we have
$$ Hom_k(V,V otimes_kW)cong Hom_k(V,V) otimes_kW$$
My attempt: EDIT ( I noticed I have typed up some nonsense)
There is a map from RHS to LHS, sending $f otimes_k w$ to element
$$ (f otimes_k w) (v) = f(v) otimes _k w $$
Motivation: this is an attempt to solve the more general question
abstract-algebra tensor-products
asked Jan 13 at 16:42
CL.CL.
2,2432825
$endgroup$
1
$begingroup$
It's not enough to show that the basis has a nontrivial image to conclude that it's an isomorphism
$endgroup$
– Max
Jan 13 at 17:48
1
$begingroup$
Moreover $f(v) (votimes w)$ makes no sense : $f(v) in V$ and $votimes win Votimes W$
$endgroup$
– Max
Jan 13 at 18:01
add a comment |
$begingroup$
I am thinking the following statement is true:
Let $V,W$ be finite dimensional $k$-vector spaces. Then we have
$$ Hom_k(V,V otimes_kW)cong Hom_k(V,V) otimes_kW$$
My attempt: EDIT ( I noticed I have typed up some nonsense)
There is a map from RHS to LHS, sending $f otimes_k w$ to element
$$ (f otimes_k w) (v) = f(v) otimes _k w $$
Motivation: this is an attempt to solve the more general question
abstract-algebra tensor-products
asked Jan 13 at 16:42
CL.CL.
2,2432825
$endgroup$
I am thinking the following statement is true:
Let $V,W$ be finite dimensional $k$-vector spaces. Then we have
$$ Hom_k(V,V otimes_kW)cong Hom_k(V,V) otimes_kW$$
My attempt: EDIT ( I noticed I have typed up some nonsense)
There is a map from RHS to LHS, sending $f otimes_k w$ to element
$$ (f otimes_k w) (v) = f(v) otimes _k w $$
Motivation: this is an attempt to solve the more general question
abstract-algebra tensor-products
abstract-algebra tensor-products
asked Jan 13 at 16:42
CL.CL.
2,2432825
asked Jan 13 at 16:42
CL.CL.
2,2432825
edited Jan 13 at 18:53
CL.
asked Jan 13 at 16:42
CL.CL.
2,2432825
asked Jan 13 at 16:42
CL.CL.
2,2432825
asked Jan 13 at 16:42
CL.CL.
2,2432825
2,2432825
1
$begingroup$
It's not enough to show that the basis has a nontrivial image to conclude that it's an isomorphism
$endgroup$
– Max
Jan 13 at 17:48
1
$begingroup$
Moreover $f(v) (votimes w)$ makes no sense : $f(v) in V$ and $votimes win Votimes W$
$endgroup$
– Max
Jan 13 at 18:01
add a comment |
1
$begingroup$
It's not enough to show that the basis has a nontrivial image to conclude that it's an isomorphism
$endgroup$
– Max
Jan 13 at 17:48
1
$begingroup$
Moreover $f(v) (votimes w)$ makes no sense : $f(v) in V$ and $votimes win Votimes W$
$endgroup$
– Max
Jan 13 at 18:01
1
1
$begingroup$
It's not enough to show that the basis has a nontrivial image to conclude that it's an isomorphism
$endgroup$
– Max
Jan 13 at 17:48
$begingroup$
It's not enough to show that the basis has a nontrivial image to conclude that it's an isomorphism
$endgroup$
– Max
Jan 13 at 17:48
1
1
$begingroup$
Moreover $f(v) (votimes w)$ makes no sense : $f(v) in V$ and $votimes win Votimes W$
$endgroup$
– Max
Jan 13 at 18:01
$begingroup$
Moreover $f(v) (votimes w)$ makes no sense : $f(v) in V$ and $votimes win Votimes W$
$endgroup$
– Max
Jan 13 at 18:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A few things : first of all, notice that they have the same dimension as $k$-vector spaces, so they are automatically isomorphic.
Now the problem is that the isomorphism you get depends a lot on $V,W$, it's not "natural" or "canonical".
Then, note that what you wrote doesn't make sense : $f(v)(votimes w)$ is not even well-typed : both $f(v)$ and $votimes w$ are vectors.
Another thing is that when you have finite dimensional vector spaces, $hom_k(E,F)simeq E^*otimes F$, the isomorphism being given from the RHS to the LHS by $lotimes v mapsto (xmapsto l(x)v)$.
So in your situation, $hom_k(V,Votimes W)simeq V^*otimes (Votimes W) simeq (V^*otimes V)otimes Wsimeq hom_k(V,V)otimes W$, which shows that for finite dimensional spaces there is actually a natural isomorphism.
Now what does my isomorphism do ? Let's see from the RHS to the LHS what it does : let $fin hom_k(V,V)$, $win W$. If the matrix of $f$ in $(e_i)$ is $(a_{ij})$ then our isomorphism sends $fotimes_k w$ to $displaystylesum_{i,j}a_{ij}(e_i^*otimes e_j otimes w)$. This is sent to $gin hom_k(V,Votimes W)$
To $v=displaystylesum_i v_i a_i$, this sends $displaystylesum_{i,j}a_{ij}e_i^*(v)(e_j otimes w) = displaystylesum_{i,j}a_{ij}v_i (e_jotimes w) = f(v)otimes w$.
Ok so now without the isomorphism I introduced we can still define $hom_k(V,V)otimes Wto hom_k(V,Votimes W), fotimes wmapsto (vmapsto f(v)otimes w)$.
To prove it is injective for instance, you can use a basis but not the way you did : it's not enough to show that your morphism is nontrivial on elements of a basis to show that it is injective. Indeed if $(e_i)$ is any basis, $displaystylesum_i e_i^*$ is nonzero on all basis elements, but it's definitely not injective in dimensions $geq 2$.
But what you can do is, calling the morphism $g$, find a suitable basis $a_i$ and see what $g(displaystylesum_i lambda_i a_i) = 0$ tells you. Here, if $(e_i), (f_j)$ are bases of $V,W$ respectively, you have a canonical basis for $hom_k (V,V)otimes W$ that can help you prove the injectivity.
answered Jan 13 at 18:23
MaxMax
14.4k11142
$endgroup$
$begingroup$
Thanks a lot. I have made severe mistake in defining the first map, I am sorry for that :( But my point wasn't in showing that the image of basis is nontrivial - but they in fact map to linearly independent elements. It seems unclear to me what you are doing at the end, but it seems it boils down to showing $Hom_k(E,F) cong E^* otimes F$? I guess this is the argument: $e_i^* otimes f_j $ maps to linearly independent elements, for a choice of basis. By dimensions, we have an isomorphism.
$endgroup$
– CL.
Jan 14 at 7:46
$begingroup$
At the end I was just showing how you could prove the map I defined is an isomorphism without using $hom(E,F)simeq E^*otimes F$. And yes, this is essentially the proof
$endgroup$
– Max
Jan 14 at 8:10
$begingroup$
I wonder what your thoughts are on the related post? I think it follows from this special case.
$endgroup$
– CL.
Jan 14 at 8:20
$begingroup$
The $Omega^i(M)$ don't seem to be finite dimensional, I don't see how it would follow (though I don't know much differential geometry). But you do have a map in one direction, perhaps you can try to see why it is an iso in your situation
$endgroup$
– Max
Jan 14 at 8:41
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A few things : first of all, notice that they have the same dimension as $k$-vector spaces, so they are automatically isomorphic.
Now the problem is that the isomorphism you get depends a lot on $V,W$, it's not "natural" or "canonical".
Then, note that what you wrote doesn't make sense : $f(v)(votimes w)$ is not even well-typed : both $f(v)$ and $votimes w$ are vectors.
Another thing is that when you have finite dimensional vector spaces, $hom_k(E,F)simeq E^*otimes F$, the isomorphism being given from the RHS to the LHS by $lotimes v mapsto (xmapsto l(x)v)$.
So in your situation, $hom_k(V,Votimes W)simeq V^*otimes (Votimes W) simeq (V^*otimes V)otimes Wsimeq hom_k(V,V)otimes W$, which shows that for finite dimensional spaces there is actually a natural isomorphism.
Now what does my isomorphism do ? Let's see from the RHS to the LHS what it does : let $fin hom_k(V,V)$, $win W$. If the matrix of $f$ in $(e_i)$ is $(a_{ij})$ then our isomorphism sends $fotimes_k w$ to $displaystylesum_{i,j}a_{ij}(e_i^*otimes e_j otimes w)$. This is sent to $gin hom_k(V,Votimes W)$
To $v=displaystylesum_i v_i a_i$, this sends $displaystylesum_{i,j}a_{ij}e_i^*(v)(e_j otimes w) = displaystylesum_{i,j}a_{ij}v_i (e_jotimes w) = f(v)otimes w$.
Ok so now without the isomorphism I introduced we can still define $hom_k(V,V)otimes Wto hom_k(V,Votimes W), fotimes wmapsto (vmapsto f(v)otimes w)$.
To prove it is injective for instance, you can use a basis but not the way you did : it's not enough to show that your morphism is nontrivial on elements of a basis to show that it is injective. Indeed if $(e_i)$ is any basis, $displaystylesum_i e_i^*$ is nonzero on all basis elements, but it's definitely not injective in dimensions $geq 2$.
But what you can do is, calling the morphism $g$, find a suitable basis $a_i$ and see what $g(displaystylesum_i lambda_i a_i) = 0$ tells you. Here, if $(e_i), (f_j)$ are bases of $V,W$ respectively, you have a canonical basis for $hom_k (V,V)otimes W$ that can help you prove the injectivity.
answered Jan 13 at 18:23
MaxMax
14.4k11142
$endgroup$
$begingroup$
Thanks a lot. I have made severe mistake in defining the first map, I am sorry for that :( But my point wasn't in showing that the image of basis is nontrivial - but they in fact map to linearly independent elements. It seems unclear to me what you are doing at the end, but it seems it boils down to showing $Hom_k(E,F) cong E^* otimes F$? I guess this is the argument: $e_i^* otimes f_j $ maps to linearly independent elements, for a choice of basis. By dimensions, we have an isomorphism.
$endgroup$
– CL.
Jan 14 at 7:46
$begingroup$
At the end I was just showing how you could prove the map I defined is an isomorphism without using $hom(E,F)simeq E^*otimes F$. And yes, this is essentially the proof
$endgroup$
– Max
Jan 14 at 8:10
$begingroup$
I wonder what your thoughts are on the related post? I think it follows from this special case.
$endgroup$
– CL.
Jan 14 at 8:20
$begingroup$
The $Omega^i(M)$ don't seem to be finite dimensional, I don't see how it would follow (though I don't know much differential geometry). But you do have a map in one direction, perhaps you can try to see why it is an iso in your situation
$endgroup$
– Max
Jan 14 at 8:41
add a comment |
$begingroup$
A few things : first of all, notice that they have the same dimension as $k$-vector spaces, so they are automatically isomorphic.
Now the problem is that the isomorphism you get depends a lot on $V,W$, it's not "natural" or "canonical".
Then, note that what you wrote doesn't make sense : $f(v)(votimes w)$ is not even well-typed : both $f(v)$ and $votimes w$ are vectors.
Another thing is that when you have finite dimensional vector spaces, $hom_k(E,F)simeq E^*otimes F$, the isomorphism being given from the RHS to the LHS by $lotimes v mapsto (xmapsto l(x)v)$.
So in your situation, $hom_k(V,Votimes W)simeq V^*otimes (Votimes W) simeq (V^*otimes V)otimes Wsimeq hom_k(V,V)otimes W$, which shows that for finite dimensional spaces there is actually a natural isomorphism.
Now what does my isomorphism do ? Let's see from the RHS to the LHS what it does : let $fin hom_k(V,V)$, $win W$. If the matrix of $f$ in $(e_i)$ is $(a_{ij})$ then our isomorphism sends $fotimes_k w$ to $displaystylesum_{i,j}a_{ij}(e_i^*otimes e_j otimes w)$. This is sent to $gin hom_k(V,Votimes W)$
To $v=displaystylesum_i v_i a_i$, this sends $displaystylesum_{i,j}a_{ij}e_i^*(v)(e_j otimes w) = displaystylesum_{i,j}a_{ij}v_i (e_jotimes w) = f(v)otimes w$.
Ok so now without the isomorphism I introduced we can still define $hom_k(V,V)otimes Wto hom_k(V,Votimes W), fotimes wmapsto (vmapsto f(v)otimes w)$.
To prove it is injective for instance, you can use a basis but not the way you did : it's not enough to show that your morphism is nontrivial on elements of a basis to show that it is injective. Indeed if $(e_i)$ is any basis, $displaystylesum_i e_i^*$ is nonzero on all basis elements, but it's definitely not injective in dimensions $geq 2$.
But what you can do is, calling the morphism $g$, find a suitable basis $a_i$ and see what $g(displaystylesum_i lambda_i a_i) = 0$ tells you. Here, if $(e_i), (f_j)$ are bases of $V,W$ respectively, you have a canonical basis for $hom_k (V,V)otimes W$ that can help you prove the injectivity.
answered Jan 13 at 18:23
MaxMax
14.4k11142
$endgroup$
$begingroup$
Thanks a lot. I have made severe mistake in defining the first map, I am sorry for that :( But my point wasn't in showing that the image of basis is nontrivial - but they in fact map to linearly independent elements. It seems unclear to me what you are doing at the end, but it seems it boils down to showing $Hom_k(E,F) cong E^* otimes F$? I guess this is the argument: $e_i^* otimes f_j $ maps to linearly independent elements, for a choice of basis. By dimensions, we have an isomorphism.
$endgroup$
– CL.
Jan 14 at 7:46
$begingroup$
At the end I was just showing how you could prove the map I defined is an isomorphism without using $hom(E,F)simeq E^*otimes F$. And yes, this is essentially the proof
$endgroup$
– Max
Jan 14 at 8:10
$begingroup$
I wonder what your thoughts are on the related post? I think it follows from this special case.
$endgroup$
– CL.
Jan 14 at 8:20
$begingroup$
The $Omega^i(M)$ don't seem to be finite dimensional, I don't see how it would follow (though I don't know much differential geometry). But you do have a map in one direction, perhaps you can try to see why it is an iso in your situation
$endgroup$
– Max
Jan 14 at 8:41
add a comment |
$begingroup$
A few things : first of all, notice that they have the same dimension as $k$-vector spaces, so they are automatically isomorphic.
Now the problem is that the isomorphism you get depends a lot on $V,W$, it's not "natural" or "canonical".
Then, note that what you wrote doesn't make sense : $f(v)(votimes w)$ is not even well-typed : both $f(v)$ and $votimes w$ are vectors.
Another thing is that when you have finite dimensional vector spaces, $hom_k(E,F)simeq E^*otimes F$, the isomorphism being given from the RHS to the LHS by $lotimes v mapsto (xmapsto l(x)v)$.
So in your situation, $hom_k(V,Votimes W)simeq V^*otimes (Votimes W) simeq (V^*otimes V)otimes Wsimeq hom_k(V,V)otimes W$, which shows that for finite dimensional spaces there is actually a natural isomorphism.
Now what does my isomorphism do ? Let's see from the RHS to the LHS what it does : let $fin hom_k(V,V)$, $win W$. If the matrix of $f$ in $(e_i)$ is $(a_{ij})$ then our isomorphism sends $fotimes_k w$ to $displaystylesum_{i,j}a_{ij}(e_i^*otimes e_j otimes w)$. This is sent to $gin hom_k(V,Votimes W)$
To $v=displaystylesum_i v_i a_i$, this sends $displaystylesum_{i,j}a_{ij}e_i^*(v)(e_j otimes w) = displaystylesum_{i,j}a_{ij}v_i (e_jotimes w) = f(v)otimes w$.
Ok so now without the isomorphism I introduced we can still define $hom_k(V,V)otimes Wto hom_k(V,Votimes W), fotimes wmapsto (vmapsto f(v)otimes w)$.
To prove it is injective for instance, you can use a basis but not the way you did : it's not enough to show that your morphism is nontrivial on elements of a basis to show that it is injective. Indeed if $(e_i)$ is any basis, $displaystylesum_i e_i^*$ is nonzero on all basis elements, but it's definitely not injective in dimensions $geq 2$.
But what you can do is, calling the morphism $g$, find a suitable basis $a_i$ and see what $g(displaystylesum_i lambda_i a_i) = 0$ tells you. Here, if $(e_i), (f_j)$ are bases of $V,W$ respectively, you have a canonical basis for $hom_k (V,V)otimes W$ that can help you prove the injectivity.
answered Jan 13 at 18:23
MaxMax
14.4k11142
$endgroup$
A few things : first of all, notice that they have the same dimension as $k$-vector spaces, so they are automatically isomorphic.
Now the problem is that the isomorphism you get depends a lot on $V,W$, it's not "natural" or "canonical".
Then, note that what you wrote doesn't make sense : $f(v)(votimes w)$ is not even well-typed : both $f(v)$ and $votimes w$ are vectors.
Another thing is that when you have finite dimensional vector spaces, $hom_k(E,F)simeq E^*otimes F$, the isomorphism being given from the RHS to the LHS by $lotimes v mapsto (xmapsto l(x)v)$.
So in your situation, $hom_k(V,Votimes W)simeq V^*otimes (Votimes W) simeq (V^*otimes V)otimes Wsimeq hom_k(V,V)otimes W$, which shows that for finite dimensional spaces there is actually a natural isomorphism.
Now what does my isomorphism do ? Let's see from the RHS to the LHS what it does : let $fin hom_k(V,V)$, $win W$. If the matrix of $f$ in $(e_i)$ is $(a_{ij})$ then our isomorphism sends $fotimes_k w$ to $displaystylesum_{i,j}a_{ij}(e_i^*otimes e_j otimes w)$. This is sent to $gin hom_k(V,Votimes W)$
To $v=displaystylesum_i v_i a_i$, this sends $displaystylesum_{i,j}a_{ij}e_i^*(v)(e_j otimes w) = displaystylesum_{i,j}a_{ij}v_i (e_jotimes w) = f(v)otimes w$.
Ok so now without the isomorphism I introduced we can still define $hom_k(V,V)otimes Wto hom_k(V,Votimes W), fotimes wmapsto (vmapsto f(v)otimes w)$.
To prove it is injective for instance, you can use a basis but not the way you did : it's not enough to show that your morphism is nontrivial on elements of a basis to show that it is injective. Indeed if $(e_i)$ is any basis, $displaystylesum_i e_i^*$ is nonzero on all basis elements, but it's definitely not injective in dimensions $geq 2$.
But what you can do is, calling the morphism $g$, find a suitable basis $a_i$ and see what $g(displaystylesum_i lambda_i a_i) = 0$ tells you. Here, if $(e_i), (f_j)$ are bases of $V,W$ respectively, you have a canonical basis for $hom_k (V,V)otimes W$ that can help you prove the injectivity.
answered Jan 13 at 18:23
MaxMax
14.4k11142
answered Jan 13 at 18:23
MaxMax
14.4k11142
answered Jan 13 at 18:23
MaxMax
14.4k11142
answered Jan 13 at 18:23
MaxMax
14.4k11142
14.4k11142
$begingroup$
Thanks a lot. I have made severe mistake in defining the first map, I am sorry for that :( But my point wasn't in showing that the image of basis is nontrivial - but they in fact map to linearly independent elements. It seems unclear to me what you are doing at the end, but it seems it boils down to showing $Hom_k(E,F) cong E^* otimes F$? I guess this is the argument: $e_i^* otimes f_j $ maps to linearly independent elements, for a choice of basis. By dimensions, we have an isomorphism.
$endgroup$
– CL.
Jan 14 at 7:46
$begingroup$
At the end I was just showing how you could prove the map I defined is an isomorphism without using $hom(E,F)simeq E^*otimes F$. And yes, this is essentially the proof
$endgroup$
– Max
Jan 14 at 8:10
$begingroup$
I wonder what your thoughts are on the related post? I think it follows from this special case.
$endgroup$
– CL.
Jan 14 at 8:20
$begingroup$
The $Omega^i(M)$ don't seem to be finite dimensional, I don't see how it would follow (though I don't know much differential geometry). But you do have a map in one direction, perhaps you can try to see why it is an iso in your situation
$endgroup$
– Max
Jan 14 at 8:41
add a comment |
$begingroup$
Thanks a lot. I have made severe mistake in defining the first map, I am sorry for that :( But my point wasn't in showing that the image of basis is nontrivial - but they in fact map to linearly independent elements. It seems unclear to me what you are doing at the end, but it seems it boils down to showing $Hom_k(E,F) cong E^* otimes F$? I guess this is the argument: $e_i^* otimes f_j $ maps to linearly independent elements, for a choice of basis. By dimensions, we have an isomorphism.
$endgroup$
– CL.
Jan 14 at 7:46
$begingroup$
At the end I was just showing how you could prove the map I defined is an isomorphism without using $hom(E,F)simeq E^*otimes F$. And yes, this is essentially the proof
$endgroup$
– Max
Jan 14 at 8:10
$begingroup$
I wonder what your thoughts are on the related post? I think it follows from this special case.
$endgroup$
– CL.
Jan 14 at 8:20
$begingroup$
The $Omega^i(M)$ don't seem to be finite dimensional, I don't see how it would follow (though I don't know much differential geometry). But you do have a map in one direction, perhaps you can try to see why it is an iso in your situation
$endgroup$
– Max
Jan 14 at 8:41
$begingroup$
Thanks a lot. I have made severe mistake in defining the first map, I am sorry for that :( But my point wasn't in showing that the image of basis is nontrivial - but they in fact map to linearly independent elements. It seems unclear to me what you are doing at the end, but it seems it boils down to showing $Hom_k(E,F) cong E^* otimes F$? I guess this is the argument: $e_i^* otimes f_j $ maps to linearly independent elements, for a choice of basis. By dimensions, we have an isomorphism.
$endgroup$
– CL.
Jan 14 at 7:46
$begingroup$
Thanks a lot. I have made severe mistake in defining the first map, I am sorry for that :( But my point wasn't in showing that the image of basis is nontrivial - but they in fact map to linearly independent elements. It seems unclear to me what you are doing at the end, but it seems it boils down to showing $Hom_k(E,F) cong E^* otimes F$? I guess this is the argument: $e_i^* otimes f_j $ maps to linearly independent elements, for a choice of basis. By dimensions, we have an isomorphism.
$endgroup$
– CL.
Jan 14 at 7:46
$begingroup$
At the end I was just showing how you could prove the map I defined is an isomorphism without using $hom(E,F)simeq E^*otimes F$. And yes, this is essentially the proof
$endgroup$
– Max
Jan 14 at 8:10
$begingroup$
At the end I was just showing how you could prove the map I defined is an isomorphism without using $hom(E,F)simeq E^*otimes F$. And yes, this is essentially the proof
$endgroup$
– Max
Jan 14 at 8:10
$begingroup$
I wonder what your thoughts are on the related post? I think it follows from this special case.
$endgroup$
– CL.
Jan 14 at 8:20
$begingroup$
I wonder what your thoughts are on the related post? I think it follows from this special case.
$endgroup$
– CL.
Jan 14 at 8:20
$begingroup$
The $Omega^i(M)$ don't seem to be finite dimensional, I don't see how it would follow (though I don't know much differential geometry). But you do have a map in one direction, perhaps you can try to see why it is an iso in your situation
$endgroup$
– Max
Jan 14 at 8:41
$begingroup$
The $Omega^i(M)$ don't seem to be finite dimensional, I don't see how it would follow (though I don't know much differential geometry). But you do have a map in one direction, perhaps you can try to see why it is an iso in your situation
$endgroup$
– Max
Jan 14 at 8:41
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1
$begingroup$
It's not enough to show that the basis has a nontrivial image to conclude that it's an isomorphism
$endgroup$
– Max
Jan 13 at 17:48
1
$begingroup$
Moreover $f(v) (votimes w)$ makes no sense : $f(v) in V$ and $votimes win Votimes W$
$endgroup$
– Max
Jan 13 at 18:01