Mixing
Dimensional Vector spoace with injective surjective and bijective
$begingroup$
I know there are similar questions on the internet, but I'm not getting it smart.
i hope u can help me,
Let $B,C$ finite dimensional Vector space with $dim(C) = dim(B)$ and $R in Hom(B,C)$.
Then it should be equivalent to R is injective, R is surjective and R is bijective.
I hope u can help me to construct a proof
linear-algebra vector-spaces injective-module
asked Jan 19 at 12:31
the mathematicthe mathematic
83
$endgroup$
add a comment |
$begingroup$
I know there are similar questions on the internet, but I'm not getting it smart.
i hope u can help me,
Let $B,C$ finite dimensional Vector space with $dim(C) = dim(B)$ and $R in Hom(B,C)$.
Then it should be equivalent to R is injective, R is surjective and R is bijective.
I hope u can help me to construct a proof
linear-algebra vector-spaces injective-module
asked Jan 19 at 12:31
the mathematicthe mathematic
83
$endgroup$
2
$begingroup$
I have no idea what you're trying to say/ask.
$endgroup$
– David C. Ullrich
Jan 19 at 12:58
1
$begingroup$
I want to construct a proof, R is injective and that's equivalent with R is surjective
$endgroup$
– the mathematic
Jan 19 at 13:12
$begingroup$
That's obviously not so - $R$ might not be injective. I really doubt that "R is injective and that's equivalent with R is surjective" is what you were asked to prove! Surely you were asked to prove that R is injective if and only if R is surjective. That's not the same as what you said! $pland(piff q)$ is not the same as $piff q$. You should state what you're asked to prove, exactly as it was given to you, without reworig anything...
$endgroup$
– David C. Ullrich
Jan 19 at 13:28
$begingroup$
sorry my mistake i changed it now
$endgroup$
– the mathematic
Jan 19 at 13:47
$begingroup$
Still makes no sense. "it should be equivalent to R is injective, R is surjective" (i) raises the question of what should be equivalent, (ii) does not say R is injective if is equivalent to R is surjective. IF you actually want to prove that R is injective if and only if R is surjective you should say "R is injective if and only if R is surjective".
$endgroup$
– David C. Ullrich
Jan 19 at 13:50
add a comment |
$begingroup$
I know there are similar questions on the internet, but I'm not getting it smart.
i hope u can help me,
Let $B,C$ finite dimensional Vector space with $dim(C) = dim(B)$ and $R in Hom(B,C)$.
Then it should be equivalent to R is injective, R is surjective and R is bijective.
I hope u can help me to construct a proof
linear-algebra vector-spaces injective-module
asked Jan 19 at 12:31
the mathematicthe mathematic
83
$endgroup$
I know there are similar questions on the internet, but I'm not getting it smart.
i hope u can help me,
Let $B,C$ finite dimensional Vector space with $dim(C) = dim(B)$ and $R in Hom(B,C)$.
Then it should be equivalent to R is injective, R is surjective and R is bijective.
I hope u can help me to construct a proof
linear-algebra vector-spaces injective-module
linear-algebra vector-spaces injective-module
asked Jan 19 at 12:31
the mathematicthe mathematic
83
asked Jan 19 at 12:31
the mathematicthe mathematic
83
edited Jan 19 at 13:46
the mathematic
asked Jan 19 at 12:31
the mathematicthe mathematic
83
asked Jan 19 at 12:31
the mathematicthe mathematic
83
asked Jan 19 at 12:31
the mathematicthe mathematic
83
83
2
$begingroup$
I have no idea what you're trying to say/ask.
$endgroup$
– David C. Ullrich
Jan 19 at 12:58
1
$begingroup$
I want to construct a proof, R is injective and that's equivalent with R is surjective
$endgroup$
– the mathematic
Jan 19 at 13:12
$begingroup$
That's obviously not so - $R$ might not be injective. I really doubt that "R is injective and that's equivalent with R is surjective" is what you were asked to prove! Surely you were asked to prove that R is injective if and only if R is surjective. That's not the same as what you said! $pland(piff q)$ is not the same as $piff q$. You should state what you're asked to prove, exactly as it was given to you, without reworig anything...
$endgroup$
– David C. Ullrich
Jan 19 at 13:28
$begingroup$
sorry my mistake i changed it now
$endgroup$
– the mathematic
Jan 19 at 13:47
$begingroup$
Still makes no sense. "it should be equivalent to R is injective, R is surjective" (i) raises the question of what should be equivalent, (ii) does not say R is injective if is equivalent to R is surjective. IF you actually want to prove that R is injective if and only if R is surjective you should say "R is injective if and only if R is surjective".
$endgroup$
– David C. Ullrich
Jan 19 at 13:50
add a comment |
2
$begingroup$
I have no idea what you're trying to say/ask.
$endgroup$
– David C. Ullrich
Jan 19 at 12:58
1
$begingroup$
I want to construct a proof, R is injective and that's equivalent with R is surjective
$endgroup$
– the mathematic
Jan 19 at 13:12
$begingroup$
That's obviously not so - $R$ might not be injective. I really doubt that "R is injective and that's equivalent with R is surjective" is what you were asked to prove! Surely you were asked to prove that R is injective if and only if R is surjective. That's not the same as what you said! $pland(piff q)$ is not the same as $piff q$. You should state what you're asked to prove, exactly as it was given to you, without reworig anything...
$endgroup$
– David C. Ullrich
Jan 19 at 13:28
$begingroup$
sorry my mistake i changed it now
$endgroup$
– the mathematic
Jan 19 at 13:47
$begingroup$
Still makes no sense. "it should be equivalent to R is injective, R is surjective" (i) raises the question of what should be equivalent, (ii) does not say R is injective if is equivalent to R is surjective. IF you actually want to prove that R is injective if and only if R is surjective you should say "R is injective if and only if R is surjective".
$endgroup$
– David C. Ullrich
Jan 19 at 13:50
2
2
$begingroup$
I have no idea what you're trying to say/ask.
$endgroup$
– David C. Ullrich
Jan 19 at 12:58
$begingroup$
I have no idea what you're trying to say/ask.
$endgroup$
– David C. Ullrich
Jan 19 at 12:58
1
1
$begingroup$
I want to construct a proof, R is injective and that's equivalent with R is surjective
$endgroup$
– the mathematic
Jan 19 at 13:12
$begingroup$
I want to construct a proof, R is injective and that's equivalent with R is surjective
$endgroup$
– the mathematic
Jan 19 at 13:12
$begingroup$
That's obviously not so - $R$ might not be injective. I really doubt that "R is injective and that's equivalent with R is surjective" is what you were asked to prove! Surely you were asked to prove that R is injective if and only if R is surjective. That's not the same as what you said! $pland(piff q)$ is not the same as $piff q$. You should state what you're asked to prove, exactly as it was given to you, without reworig anything...
$endgroup$
– David C. Ullrich
Jan 19 at 13:28
$begingroup$
That's obviously not so - $R$ might not be injective. I really doubt that "R is injective and that's equivalent with R is surjective" is what you were asked to prove! Surely you were asked to prove that R is injective if and only if R is surjective. That's not the same as what you said! $pland(piff q)$ is not the same as $piff q$. You should state what you're asked to prove, exactly as it was given to you, without reworig anything...
$endgroup$
– David C. Ullrich
Jan 19 at 13:28
$begingroup$
sorry my mistake i changed it now
$endgroup$
– the mathematic
Jan 19 at 13:47
$begingroup$
sorry my mistake i changed it now
$endgroup$
– the mathematic
Jan 19 at 13:47
$begingroup$
Still makes no sense. "it should be equivalent to R is injective, R is surjective" (i) raises the question of what should be equivalent, (ii) does not say R is injective if is equivalent to R is surjective. IF you actually want to prove that R is injective if and only if R is surjective you should say "R is injective if and only if R is surjective".
$endgroup$
– David C. Ullrich
Jan 19 at 13:50
$begingroup$
Still makes no sense. "it should be equivalent to R is injective, R is surjective" (i) raises the question of what should be equivalent, (ii) does not say R is injective if is equivalent to R is surjective. IF you actually want to prove that R is injective if and only if R is surjective you should say "R is injective if and only if R is surjective".
$endgroup$
– David C. Ullrich
Jan 19 at 13:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As @David has mentioned, it would help if you made the question clearer. Since the statement does not hold for infinite-dimensional vector spaces, is it possible that you actually want to prove it for finite-dimensional vector spaces?
If that is the case, choose a basis for $B$; say, $lbrace x_1,dots ,x_nrbrace$. If $R$ is injective, try showing that $lbrace R(x_1),dots ,R(x_n)rbrace$ is linearly independent in $C$. Here you must use (a) that $R$ is injective and (b) that $R$ is linear. Then conclude from dimension considerations that $mathrm{span} (R(x_1),dots ,R(x_n))$ is a basis for $C$, hence showing that $R$ is surjective.
The proof of surjective $Rightarrow$ injective is somewhat similar.
Walkthrough. In case you give up, here is a walkthrough of the proof for injective $Rightarrow$ surjective. Let $lbrace x_1,dots ,x_nrbrace$ be as above. Then $mathrm{dim} (B) = mathrm{dim} (C) = n$. We will show that $lbrace R(x_1),dots ,R(x_n)rbrace$ is a basis for $C$.
First we must see that $R(x_1),dots ,R(x_n)$ are linearly independent. Suppose we have
$$
alpha_1R(x_1)+dots +alpha_n R(x_n) = 0.
$$
But $R$ is linear, so we get:
$$
R(alpha_1x_1 + dots + alpha_nx_n) = 0.
$$
Since $R$ is injective, we then have
$$
alpha_1x_1 + dots + alpha_n x_n = 0.
$$
But $x_1,dots ,x_n$ are linearly independent, so we must have $alpha_i = 0$ for each $i=1,dots ,n$. This shows that $R(x_1),dots ,R(x_n)$ are linearly independent. Hence they must span an $n$-dimensional subspace of $C$. Since $C$ is $n$-dimensional, this must be all of $C$. But then $R$ is surjective.
answered Jan 19 at 13:41
o.h.o.h.
4616
$endgroup$
$begingroup$
I was hoping to redirect him towards something more fruitful. That's why I wrote "is it possible that you actually want to prove it for finite-dimensional vector spaces?". I think it's likely that he's misread what he has to prove.
$endgroup$
– o.h.
Jan 19 at 13:46
$begingroup$
yeah thank you, i was indeed misreading
$endgroup$
– the mathematic
Jan 19 at 13:50
add a comment |
$begingroup$
Note: This is a reply to the original version of the question, where it was specified that $B$ and $C$ were "endless dimensional". So of course it's irrelevant to the current version, where $B$ and $C$ are finite-dimensional. May as well leave it here, to show that the hypothesis is needed.
The question is totally unclear, and the comment doesn't clarify things much. It seems possible that you actually want to prove that $R$ is injective if and only if $R$ is surjective. If so:
(i) you should say so!
(ii) you can't prove that because it's false. Say $B$ is the space of sequences of real numbers. Let $C=B$, and define $R:Bto C$ by $$Rx=(0,x_1,x_2,dots).$$
Then $R$ is injective but not surjective.
You should give a similar example of a linear map which is surjective but not injective.
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As @David has mentioned, it would help if you made the question clearer. Since the statement does not hold for infinite-dimensional vector spaces, is it possible that you actually want to prove it for finite-dimensional vector spaces?
If that is the case, choose a basis for $B$; say, $lbrace x_1,dots ,x_nrbrace$. If $R$ is injective, try showing that $lbrace R(x_1),dots ,R(x_n)rbrace$ is linearly independent in $C$. Here you must use (a) that $R$ is injective and (b) that $R$ is linear. Then conclude from dimension considerations that $mathrm{span} (R(x_1),dots ,R(x_n))$ is a basis for $C$, hence showing that $R$ is surjective.
The proof of surjective $Rightarrow$ injective is somewhat similar.
Walkthrough. In case you give up, here is a walkthrough of the proof for injective $Rightarrow$ surjective. Let $lbrace x_1,dots ,x_nrbrace$ be as above. Then $mathrm{dim} (B) = mathrm{dim} (C) = n$. We will show that $lbrace R(x_1),dots ,R(x_n)rbrace$ is a basis for $C$.
First we must see that $R(x_1),dots ,R(x_n)$ are linearly independent. Suppose we have
$$
alpha_1R(x_1)+dots +alpha_n R(x_n) = 0.
$$
But $R$ is linear, so we get:
$$
R(alpha_1x_1 + dots + alpha_nx_n) = 0.
$$
Since $R$ is injective, we then have
$$
alpha_1x_1 + dots + alpha_n x_n = 0.
$$
But $x_1,dots ,x_n$ are linearly independent, so we must have $alpha_i = 0$ for each $i=1,dots ,n$. This shows that $R(x_1),dots ,R(x_n)$ are linearly independent. Hence they must span an $n$-dimensional subspace of $C$. Since $C$ is $n$-dimensional, this must be all of $C$. But then $R$ is surjective.
answered Jan 19 at 13:41
o.h.o.h.
4616
$endgroup$
$begingroup$
I was hoping to redirect him towards something more fruitful. That's why I wrote "is it possible that you actually want to prove it for finite-dimensional vector spaces?". I think it's likely that he's misread what he has to prove.
$endgroup$
– o.h.
Jan 19 at 13:46
$begingroup$
yeah thank you, i was indeed misreading
$endgroup$
– the mathematic
Jan 19 at 13:50
add a comment |
$begingroup$
As @David has mentioned, it would help if you made the question clearer. Since the statement does not hold for infinite-dimensional vector spaces, is it possible that you actually want to prove it for finite-dimensional vector spaces?
If that is the case, choose a basis for $B$; say, $lbrace x_1,dots ,x_nrbrace$. If $R$ is injective, try showing that $lbrace R(x_1),dots ,R(x_n)rbrace$ is linearly independent in $C$. Here you must use (a) that $R$ is injective and (b) that $R$ is linear. Then conclude from dimension considerations that $mathrm{span} (R(x_1),dots ,R(x_n))$ is a basis for $C$, hence showing that $R$ is surjective.
The proof of surjective $Rightarrow$ injective is somewhat similar.
Walkthrough. In case you give up, here is a walkthrough of the proof for injective $Rightarrow$ surjective. Let $lbrace x_1,dots ,x_nrbrace$ be as above. Then $mathrm{dim} (B) = mathrm{dim} (C) = n$. We will show that $lbrace R(x_1),dots ,R(x_n)rbrace$ is a basis for $C$.
First we must see that $R(x_1),dots ,R(x_n)$ are linearly independent. Suppose we have
$$
alpha_1R(x_1)+dots +alpha_n R(x_n) = 0.
$$
But $R$ is linear, so we get:
$$
R(alpha_1x_1 + dots + alpha_nx_n) = 0.
$$
Since $R$ is injective, we then have
$$
alpha_1x_1 + dots + alpha_n x_n = 0.
$$
But $x_1,dots ,x_n$ are linearly independent, so we must have $alpha_i = 0$ for each $i=1,dots ,n$. This shows that $R(x_1),dots ,R(x_n)$ are linearly independent. Hence they must span an $n$-dimensional subspace of $C$. Since $C$ is $n$-dimensional, this must be all of $C$. But then $R$ is surjective.
answered Jan 19 at 13:41
o.h.o.h.
4616
$endgroup$
$begingroup$
I was hoping to redirect him towards something more fruitful. That's why I wrote "is it possible that you actually want to prove it for finite-dimensional vector spaces?". I think it's likely that he's misread what he has to prove.
$endgroup$
– o.h.
Jan 19 at 13:46
$begingroup$
yeah thank you, i was indeed misreading
$endgroup$
– the mathematic
Jan 19 at 13:50
add a comment |
$begingroup$
As @David has mentioned, it would help if you made the question clearer. Since the statement does not hold for infinite-dimensional vector spaces, is it possible that you actually want to prove it for finite-dimensional vector spaces?
If that is the case, choose a basis for $B$; say, $lbrace x_1,dots ,x_nrbrace$. If $R$ is injective, try showing that $lbrace R(x_1),dots ,R(x_n)rbrace$ is linearly independent in $C$. Here you must use (a) that $R$ is injective and (b) that $R$ is linear. Then conclude from dimension considerations that $mathrm{span} (R(x_1),dots ,R(x_n))$ is a basis for $C$, hence showing that $R$ is surjective.
The proof of surjective $Rightarrow$ injective is somewhat similar.
Walkthrough. In case you give up, here is a walkthrough of the proof for injective $Rightarrow$ surjective. Let $lbrace x_1,dots ,x_nrbrace$ be as above. Then $mathrm{dim} (B) = mathrm{dim} (C) = n$. We will show that $lbrace R(x_1),dots ,R(x_n)rbrace$ is a basis for $C$.
First we must see that $R(x_1),dots ,R(x_n)$ are linearly independent. Suppose we have
$$
alpha_1R(x_1)+dots +alpha_n R(x_n) = 0.
$$
But $R$ is linear, so we get:
$$
R(alpha_1x_1 + dots + alpha_nx_n) = 0.
$$
Since $R$ is injective, we then have
$$
alpha_1x_1 + dots + alpha_n x_n = 0.
$$
But $x_1,dots ,x_n$ are linearly independent, so we must have $alpha_i = 0$ for each $i=1,dots ,n$. This shows that $R(x_1),dots ,R(x_n)$ are linearly independent. Hence they must span an $n$-dimensional subspace of $C$. Since $C$ is $n$-dimensional, this must be all of $C$. But then $R$ is surjective.
answered Jan 19 at 13:41
o.h.o.h.
4616
$endgroup$
As @David has mentioned, it would help if you made the question clearer. Since the statement does not hold for infinite-dimensional vector spaces, is it possible that you actually want to prove it for finite-dimensional vector spaces?
If that is the case, choose a basis for $B$; say, $lbrace x_1,dots ,x_nrbrace$. If $R$ is injective, try showing that $lbrace R(x_1),dots ,R(x_n)rbrace$ is linearly independent in $C$. Here you must use (a) that $R$ is injective and (b) that $R$ is linear. Then conclude from dimension considerations that $mathrm{span} (R(x_1),dots ,R(x_n))$ is a basis for $C$, hence showing that $R$ is surjective.
The proof of surjective $Rightarrow$ injective is somewhat similar.
Walkthrough. In case you give up, here is a walkthrough of the proof for injective $Rightarrow$ surjective. Let $lbrace x_1,dots ,x_nrbrace$ be as above. Then $mathrm{dim} (B) = mathrm{dim} (C) = n$. We will show that $lbrace R(x_1),dots ,R(x_n)rbrace$ is a basis for $C$.
First we must see that $R(x_1),dots ,R(x_n)$ are linearly independent. Suppose we have
$$
alpha_1R(x_1)+dots +alpha_n R(x_n) = 0.
$$
But $R$ is linear, so we get:
$$
R(alpha_1x_1 + dots + alpha_nx_n) = 0.
$$
Since $R$ is injective, we then have
$$
alpha_1x_1 + dots + alpha_n x_n = 0.
$$
But $x_1,dots ,x_n$ are linearly independent, so we must have $alpha_i = 0$ for each $i=1,dots ,n$. This shows that $R(x_1),dots ,R(x_n)$ are linearly independent. Hence they must span an $n$-dimensional subspace of $C$. Since $C$ is $n$-dimensional, this must be all of $C$. But then $R$ is surjective.
answered Jan 19 at 13:41
o.h.o.h.
4616
edited Jan 19 at 13:58
answered Jan 19 at 13:41
o.h.o.h.
4616
answered Jan 19 at 13:41
o.h.o.h.
4616
answered Jan 19 at 13:41
o.h.o.h.
4616
4616
$begingroup$
I was hoping to redirect him towards something more fruitful. That's why I wrote "is it possible that you actually want to prove it for finite-dimensional vector spaces?". I think it's likely that he's misread what he has to prove.
$endgroup$
– o.h.
Jan 19 at 13:46
$begingroup$
yeah thank you, i was indeed misreading
$endgroup$
– the mathematic
Jan 19 at 13:50
add a comment |
$begingroup$
I was hoping to redirect him towards something more fruitful. That's why I wrote "is it possible that you actually want to prove it for finite-dimensional vector spaces?". I think it's likely that he's misread what he has to prove.
$endgroup$
– o.h.
Jan 19 at 13:46
$begingroup$
yeah thank you, i was indeed misreading
$endgroup$
– the mathematic
Jan 19 at 13:50
$begingroup$
I was hoping to redirect him towards something more fruitful. That's why I wrote "is it possible that you actually want to prove it for finite-dimensional vector spaces?". I think it's likely that he's misread what he has to prove.
$endgroup$
– o.h.
Jan 19 at 13:46
$begingroup$
I was hoping to redirect him towards something more fruitful. That's why I wrote "is it possible that you actually want to prove it for finite-dimensional vector spaces?". I think it's likely that he's misread what he has to prove.
$endgroup$
– o.h.
Jan 19 at 13:46
$begingroup$
yeah thank you, i was indeed misreading
$endgroup$
– the mathematic
Jan 19 at 13:50
$begingroup$
yeah thank you, i was indeed misreading
$endgroup$
– the mathematic
Jan 19 at 13:50
add a comment |
$begingroup$
Note: This is a reply to the original version of the question, where it was specified that $B$ and $C$ were "endless dimensional". So of course it's irrelevant to the current version, where $B$ and $C$ are finite-dimensional. May as well leave it here, to show that the hypothesis is needed.
The question is totally unclear, and the comment doesn't clarify things much. It seems possible that you actually want to prove that $R$ is injective if and only if $R$ is surjective. If so:
(i) you should say so!
(ii) you can't prove that because it's false. Say $B$ is the space of sequences of real numbers. Let $C=B$, and define $R:Bto C$ by $$Rx=(0,x_1,x_2,dots).$$
Then $R$ is injective but not surjective.
You should give a similar example of a linear map which is surjective but not injective.
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
Note: This is a reply to the original version of the question, where it was specified that $B$ and $C$ were "endless dimensional". So of course it's irrelevant to the current version, where $B$ and $C$ are finite-dimensional. May as well leave it here, to show that the hypothesis is needed.
The question is totally unclear, and the comment doesn't clarify things much. It seems possible that you actually want to prove that $R$ is injective if and only if $R$ is surjective. If so:
(i) you should say so!
(ii) you can't prove that because it's false. Say $B$ is the space of sequences of real numbers. Let $C=B$, and define $R:Bto C$ by $$Rx=(0,x_1,x_2,dots).$$
Then $R$ is injective but not surjective.
You should give a similar example of a linear map which is surjective but not injective.
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
$endgroup$
add a comment |
$begingroup$
Note: This is a reply to the original version of the question, where it was specified that $B$ and $C$ were "endless dimensional". So of course it's irrelevant to the current version, where $B$ and $C$ are finite-dimensional. May as well leave it here, to show that the hypothesis is needed.
The question is totally unclear, and the comment doesn't clarify things much. It seems possible that you actually want to prove that $R$ is injective if and only if $R$ is surjective. If so:
(i) you should say so!
(ii) you can't prove that because it's false. Say $B$ is the space of sequences of real numbers. Let $C=B$, and define $R:Bto C$ by $$Rx=(0,x_1,x_2,dots).$$
Then $R$ is injective but not surjective.
You should give a similar example of a linear map which is surjective but not injective.
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
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Note: This is a reply to the original version of the question, where it was specified that $B$ and $C$ were "endless dimensional". So of course it's irrelevant to the current version, where $B$ and $C$ are finite-dimensional. May as well leave it here, to show that the hypothesis is needed.
The question is totally unclear, and the comment doesn't clarify things much. It seems possible that you actually want to prove that $R$ is injective if and only if $R$ is surjective. If so:
(i) you should say so!
(ii) you can't prove that because it's false. Say $B$ is the space of sequences of real numbers. Let $C=B$, and define $R:Bto C$ by $$Rx=(0,x_1,x_2,dots).$$
Then $R$ is injective but not surjective.
You should give a similar example of a linear map which is surjective but not injective.
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
edited Jan 19 at 14:07
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
answered Jan 19 at 13:32
David C. UllrichDavid C. Ullrich
61k43994
61k43994
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I have no idea what you're trying to say/ask.
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– David C. Ullrich
Jan 19 at 12:58
1
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I want to construct a proof, R is injective and that's equivalent with R is surjective
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– the mathematic
Jan 19 at 13:12
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That's obviously not so - $R$ might not be injective. I really doubt that "R is injective and that's equivalent with R is surjective" is what you were asked to prove! Surely you were asked to prove that R is injective if and only if R is surjective. That's not the same as what you said! $pland(piff q)$ is not the same as $piff q$. You should state what you're asked to prove, exactly as it was given to you, without reworig anything...
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– David C. Ullrich
Jan 19 at 13:28
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sorry my mistake i changed it now
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– the mathematic
Jan 19 at 13:47
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Still makes no sense. "it should be equivalent to R is injective, R is surjective" (i) raises the question of what should be equivalent, (ii) does not say R is injective if is equivalent to R is surjective. IF you actually want to prove that R is injective if and only if R is surjective you should say "R is injective if and only if R is surjective".
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– David C. Ullrich
Jan 19 at 13:50