Mixing
Proving that if |domain (f)|= |codomain (f)| where f:A→B is surjective, it will be injective too
$begingroup$
I thought about going to prove this by contradiction.
Assume that the domain and the codomain are finite.
Assuming for contradiction that if f(a)=f(b);a≠b;a,b∈A
Because of |A|=|B|, A and B are finite, there exists e∈B for which there does not exist d∈A: f(d)=e
Therefore f is not surjective which leads us to the contradiction. So f must be injective too.
The issue is that I am not sure if this proof is correct formulated as I am a beginner. So I would like to see opinions or ways about how can I fix this, thanks.
abstract-algebra functions
edited Jan 13 at 18:47
jgon
14.5k22042
asked Jan 13 at 18:17
ValVal
557
$endgroup$
add a comment |
$begingroup$
I thought about going to prove this by contradiction.
Assume that the domain and the codomain are finite.
Assuming for contradiction that if f(a)=f(b);a≠b;a,b∈A
Because of |A|=|B|, A and B are finite, there exists e∈B for which there does not exist d∈A: f(d)=e
Therefore f is not surjective which leads us to the contradiction. So f must be injective too.
The issue is that I am not sure if this proof is correct formulated as I am a beginner. So I would like to see opinions or ways about how can I fix this, thanks.
abstract-algebra functions
edited Jan 13 at 18:47
jgon
14.5k22042
asked Jan 13 at 18:17
ValVal
557
$endgroup$
2
$begingroup$
There's a fairly major assumption missing from the question: it's important that both the domain and the codomain are finite, otherwise it's false (consider the function from the integers to the naturals sending every $x$ to $2x$, for example: how would your proof break down here? If it's not clear, change your proof until it is)
$endgroup$
– user3482749
Jan 13 at 18:20
$begingroup$
I have edited, is it now a proper proof @user3482749?
$endgroup$
– Val
Jan 13 at 18:30
$begingroup$
Notice that adding that didn't actually change your proof in any way: you haven't used the finiteness anywhere, or rather you haven't said where you're using it. Which step in your proof uses the finiteness?
$endgroup$
– user3482749
Jan 13 at 18:32
$begingroup$
I mentioned that A and B are finite now @user3482749
$endgroup$
– Val
Jan 13 at 18:47
add a comment |
$begingroup$
I thought about going to prove this by contradiction.
Assume that the domain and the codomain are finite.
Assuming for contradiction that if f(a)=f(b);a≠b;a,b∈A
Because of |A|=|B|, A and B are finite, there exists e∈B for which there does not exist d∈A: f(d)=e
Therefore f is not surjective which leads us to the contradiction. So f must be injective too.
The issue is that I am not sure if this proof is correct formulated as I am a beginner. So I would like to see opinions or ways about how can I fix this, thanks.
abstract-algebra functions
edited Jan 13 at 18:47
jgon
14.5k22042
asked Jan 13 at 18:17
ValVal
557
$endgroup$
I thought about going to prove this by contradiction.
Assume that the domain and the codomain are finite.
Assuming for contradiction that if f(a)=f(b);a≠b;a,b∈A
Because of |A|=|B|, A and B are finite, there exists e∈B for which there does not exist d∈A: f(d)=e
Therefore f is not surjective which leads us to the contradiction. So f must be injective too.
The issue is that I am not sure if this proof is correct formulated as I am a beginner. So I would like to see opinions or ways about how can I fix this, thanks.
abstract-algebra functions
abstract-algebra functions
edited Jan 13 at 18:47
jgon
14.5k22042
asked Jan 13 at 18:17
ValVal
557
edited Jan 13 at 18:47
jgon
14.5k22042
asked Jan 13 at 18:17
ValVal
557
edited Jan 13 at 18:47
jgon
14.5k22042
edited Jan 13 at 18:47
jgon
14.5k22042
edited Jan 13 at 18:47
jgon
14.5k22042
14.5k22042
asked Jan 13 at 18:17
ValVal
557
asked Jan 13 at 18:17
ValVal
557
asked Jan 13 at 18:17
ValVal
557
557
2
$begingroup$
There's a fairly major assumption missing from the question: it's important that both the domain and the codomain are finite, otherwise it's false (consider the function from the integers to the naturals sending every $x$ to $2x$, for example: how would your proof break down here? If it's not clear, change your proof until it is)
$endgroup$
– user3482749
Jan 13 at 18:20
$begingroup$
I have edited, is it now a proper proof @user3482749?
$endgroup$
– Val
Jan 13 at 18:30
$begingroup$
Notice that adding that didn't actually change your proof in any way: you haven't used the finiteness anywhere, or rather you haven't said where you're using it. Which step in your proof uses the finiteness?
$endgroup$
– user3482749
Jan 13 at 18:32
$begingroup$
I mentioned that A and B are finite now @user3482749
$endgroup$
– Val
Jan 13 at 18:47
add a comment |
2
$begingroup$
There's a fairly major assumption missing from the question: it's important that both the domain and the codomain are finite, otherwise it's false (consider the function from the integers to the naturals sending every $x$ to $2x$, for example: how would your proof break down here? If it's not clear, change your proof until it is)
$endgroup$
– user3482749
Jan 13 at 18:20
$begingroup$
I have edited, is it now a proper proof @user3482749?
$endgroup$
– Val
Jan 13 at 18:30
$begingroup$
Notice that adding that didn't actually change your proof in any way: you haven't used the finiteness anywhere, or rather you haven't said where you're using it. Which step in your proof uses the finiteness?
$endgroup$
– user3482749
Jan 13 at 18:32
$begingroup$
I mentioned that A and B are finite now @user3482749
$endgroup$
– Val
Jan 13 at 18:47
2
2
$begingroup$
There's a fairly major assumption missing from the question: it's important that both the domain and the codomain are finite, otherwise it's false (consider the function from the integers to the naturals sending every $x$ to $2x$, for example: how would your proof break down here? If it's not clear, change your proof until it is)
$endgroup$
– user3482749
Jan 13 at 18:20
$begingroup$
There's a fairly major assumption missing from the question: it's important that both the domain and the codomain are finite, otherwise it's false (consider the function from the integers to the naturals sending every $x$ to $2x$, for example: how would your proof break down here? If it's not clear, change your proof until it is)
$endgroup$
– user3482749
Jan 13 at 18:20
$begingroup$
I have edited, is it now a proper proof @user3482749?
$endgroup$
– Val
Jan 13 at 18:30
$begingroup$
I have edited, is it now a proper proof @user3482749?
$endgroup$
– Val
Jan 13 at 18:30
$begingroup$
Notice that adding that didn't actually change your proof in any way: you haven't used the finiteness anywhere, or rather you haven't said where you're using it. Which step in your proof uses the finiteness?
$endgroup$
– user3482749
Jan 13 at 18:32
$begingroup$
Notice that adding that didn't actually change your proof in any way: you haven't used the finiteness anywhere, or rather you haven't said where you're using it. Which step in your proof uses the finiteness?
$endgroup$
– user3482749
Jan 13 at 18:32
$begingroup$
I mentioned that A and B are finite now @user3482749
$endgroup$
– Val
Jan 13 at 18:47
$begingroup$
I mentioned that A and B are finite now @user3482749
$endgroup$
– Val
Jan 13 at 18:47
add a comment |
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$begingroup$
There's a fairly major assumption missing from the question: it's important that both the domain and the codomain are finite, otherwise it's false (consider the function from the integers to the naturals sending every $x$ to $2x$, for example: how would your proof break down here? If it's not clear, change your proof until it is)
$endgroup$
– user3482749
Jan 13 at 18:20
$begingroup$
I have edited, is it now a proper proof @user3482749?
$endgroup$
– Val
Jan 13 at 18:30
$begingroup$
Notice that adding that didn't actually change your proof in any way: you haven't used the finiteness anywhere, or rather you haven't said where you're using it. Which step in your proof uses the finiteness?
$endgroup$
– user3482749
Jan 13 at 18:32
$begingroup$
I mentioned that A and B are finite now @user3482749
$endgroup$
– Val
Jan 13 at 18:47