Mixing
Can an element with no pre-image be the pre-image of some element?
$begingroup$
This might be something absolutely elementary, ridiculous or obvious (or way too complex) but the thing is, I have no clue and have found no information about it.
Let's say we have some set resulting from a mapping, and that set contains some element e which has no pre-image in the domain of the function, now we compose this function with one another such that this set defines itself a domain. How can this element e be the pre-image of some element in the new codomain if it has no inverse operator? Isn't it that we do not have enough information about it as to act upon it? I somehow get the idea that in mathematics the behaviour of things is far more important than their presence.
Plus, can somebody recommend me some book or source which will explain elementary stuff like this such that people which really are looking for just elementary stuff will understand it?
linear-algebra functions linear-transformations
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
$endgroup$
add a comment |
$begingroup$
This might be something absolutely elementary, ridiculous or obvious (or way too complex) but the thing is, I have no clue and have found no information about it.
Let's say we have some set resulting from a mapping, and that set contains some element e which has no pre-image in the domain of the function, now we compose this function with one another such that this set defines itself a domain. How can this element e be the pre-image of some element in the new codomain if it has no inverse operator? Isn't it that we do not have enough information about it as to act upon it? I somehow get the idea that in mathematics the behaviour of things is far more important than their presence.
Plus, can somebody recommend me some book or source which will explain elementary stuff like this such that people which really are looking for just elementary stuff will understand it?
linear-algebra functions linear-transformations
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
$endgroup$
2
$begingroup$
I'm a little confused about the question. You ask us to consider "some set resulting from a mapping, and that set contains some element e which has no pre-image", but how could $e$ be in such a set if it has no pre-image? Could you perhaps give a simple example?
$endgroup$
– Theo C.
Feb 1 at 21:56
$begingroup$
Oh it might be that 'resulting' is not the right expression, I rather meant.. a function which maps some elements onto another set so that there are elements in that set which have no pre-image (can't be obtained with the function?). It is absolutely possible that I am confusing many concepts and have misunderstood injective functions.
$endgroup$
– houda el fezzak
Feb 1 at 22:03
add a comment |
$begingroup$
This might be something absolutely elementary, ridiculous or obvious (or way too complex) but the thing is, I have no clue and have found no information about it.
Let's say we have some set resulting from a mapping, and that set contains some element e which has no pre-image in the domain of the function, now we compose this function with one another such that this set defines itself a domain. How can this element e be the pre-image of some element in the new codomain if it has no inverse operator? Isn't it that we do not have enough information about it as to act upon it? I somehow get the idea that in mathematics the behaviour of things is far more important than their presence.
Plus, can somebody recommend me some book or source which will explain elementary stuff like this such that people which really are looking for just elementary stuff will understand it?
linear-algebra functions linear-transformations
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
$endgroup$
This might be something absolutely elementary, ridiculous or obvious (or way too complex) but the thing is, I have no clue and have found no information about it.
Let's say we have some set resulting from a mapping, and that set contains some element e which has no pre-image in the domain of the function, now we compose this function with one another such that this set defines itself a domain. How can this element e be the pre-image of some element in the new codomain if it has no inverse operator? Isn't it that we do not have enough information about it as to act upon it? I somehow get the idea that in mathematics the behaviour of things is far more important than their presence.
Plus, can somebody recommend me some book or source which will explain elementary stuff like this such that people which really are looking for just elementary stuff will understand it?
linear-algebra functions linear-transformations
linear-algebra functions linear-transformations
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
asked Feb 1 at 21:51
houda el fezzakhouda el fezzak
274
274
2
$begingroup$
I'm a little confused about the question. You ask us to consider "some set resulting from a mapping, and that set contains some element e which has no pre-image", but how could $e$ be in such a set if it has no pre-image? Could you perhaps give a simple example?
$endgroup$
– Theo C.
Feb 1 at 21:56
$begingroup$
Oh it might be that 'resulting' is not the right expression, I rather meant.. a function which maps some elements onto another set so that there are elements in that set which have no pre-image (can't be obtained with the function?). It is absolutely possible that I am confusing many concepts and have misunderstood injective functions.
$endgroup$
– houda el fezzak
Feb 1 at 22:03
add a comment |
2
$begingroup$
I'm a little confused about the question. You ask us to consider "some set resulting from a mapping, and that set contains some element e which has no pre-image", but how could $e$ be in such a set if it has no pre-image? Could you perhaps give a simple example?
$endgroup$
– Theo C.
Feb 1 at 21:56
$begingroup$
Oh it might be that 'resulting' is not the right expression, I rather meant.. a function which maps some elements onto another set so that there are elements in that set which have no pre-image (can't be obtained with the function?). It is absolutely possible that I am confusing many concepts and have misunderstood injective functions.
$endgroup$
– houda el fezzak
Feb 1 at 22:03
2
2
$begingroup$
I'm a little confused about the question. You ask us to consider "some set resulting from a mapping, and that set contains some element e which has no pre-image", but how could $e$ be in such a set if it has no pre-image? Could you perhaps give a simple example?
$endgroup$
– Theo C.
Feb 1 at 21:56
$begingroup$
I'm a little confused about the question. You ask us to consider "some set resulting from a mapping, and that set contains some element e which has no pre-image", but how could $e$ be in such a set if it has no pre-image? Could you perhaps give a simple example?
$endgroup$
– Theo C.
Feb 1 at 21:56
$begingroup$
Oh it might be that 'resulting' is not the right expression, I rather meant.. a function which maps some elements onto another set so that there are elements in that set which have no pre-image (can't be obtained with the function?). It is absolutely possible that I am confusing many concepts and have misunderstood injective functions.
$endgroup$
– houda el fezzak
Feb 1 at 22:03
$begingroup$
Oh it might be that 'resulting' is not the right expression, I rather meant.. a function which maps some elements onto another set so that there are elements in that set which have no pre-image (can't be obtained with the function?). It is absolutely possible that I am confusing many concepts and have misunderstood injective functions.
$endgroup$
– houda el fezzak
Feb 1 at 22:03
add a comment |
1 Answer
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$begingroup$
If I understand what you're asking then consider two functions:
$f: A rightarrow B, f(x)=x^2$ where $A = B = mathbb{R}$
$g: Xrightarrow Y, g(x)=x^2$ where $X = mathbb{R}$ and $Y = mathbb{R}_{geq 0}$ (real numbers $geq 0$).
Note that $-1 in A$ and $-1 in B$, but certainly there is no real number $r$ such that $r^2=-1$. As such, we say that $f$ is not onto. However, note that we still have $f(-1) = 1$, so $-1$ is an element with no pre-image but is itself the preimage of something.
Now, consider $g$ and note that $forall y in Y$, there is some $xin X$ such that $g(x)=y$, namely $x=pmsqrt{y}$. Since $g$ can map to any element of $Y$ (the co-domain), we call $g$ an "onto" or "surjective" map. Notice that simply by changing our codomain the mapping $xmapsto x^2$ can be surjective. As such, it's often very important to specify domains and codomains along with mappings when talking about functions.
Hopefully this cleared some things up!
answered Feb 1 at 22:15
Theo C.Theo C.
44228
$endgroup$
1
$begingroup$
It did thanks! I completely forgot that we are free to restrict our codomain!
$endgroup$
– houda el fezzak
Feb 3 at 21:34
add a comment |
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1 Answer
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$begingroup$
If I understand what you're asking then consider two functions:
$f: A rightarrow B, f(x)=x^2$ where $A = B = mathbb{R}$
$g: Xrightarrow Y, g(x)=x^2$ where $X = mathbb{R}$ and $Y = mathbb{R}_{geq 0}$ (real numbers $geq 0$).
Note that $-1 in A$ and $-1 in B$, but certainly there is no real number $r$ such that $r^2=-1$. As such, we say that $f$ is not onto. However, note that we still have $f(-1) = 1$, so $-1$ is an element with no pre-image but is itself the preimage of something.
Now, consider $g$ and note that $forall y in Y$, there is some $xin X$ such that $g(x)=y$, namely $x=pmsqrt{y}$. Since $g$ can map to any element of $Y$ (the co-domain), we call $g$ an "onto" or "surjective" map. Notice that simply by changing our codomain the mapping $xmapsto x^2$ can be surjective. As such, it's often very important to specify domains and codomains along with mappings when talking about functions.
Hopefully this cleared some things up!
answered Feb 1 at 22:15
Theo C.Theo C.
44228
$endgroup$
1
$begingroup$
It did thanks! I completely forgot that we are free to restrict our codomain!
$endgroup$
– houda el fezzak
Feb 3 at 21:34
add a comment |
$begingroup$
If I understand what you're asking then consider two functions:
$f: A rightarrow B, f(x)=x^2$ where $A = B = mathbb{R}$
$g: Xrightarrow Y, g(x)=x^2$ where $X = mathbb{R}$ and $Y = mathbb{R}_{geq 0}$ (real numbers $geq 0$).
Note that $-1 in A$ and $-1 in B$, but certainly there is no real number $r$ such that $r^2=-1$. As such, we say that $f$ is not onto. However, note that we still have $f(-1) = 1$, so $-1$ is an element with no pre-image but is itself the preimage of something.
Now, consider $g$ and note that $forall y in Y$, there is some $xin X$ such that $g(x)=y$, namely $x=pmsqrt{y}$. Since $g$ can map to any element of $Y$ (the co-domain), we call $g$ an "onto" or "surjective" map. Notice that simply by changing our codomain the mapping $xmapsto x^2$ can be surjective. As such, it's often very important to specify domains and codomains along with mappings when talking about functions.
Hopefully this cleared some things up!
answered Feb 1 at 22:15
Theo C.Theo C.
44228
$endgroup$
1
$begingroup$
It did thanks! I completely forgot that we are free to restrict our codomain!
$endgroup$
– houda el fezzak
Feb 3 at 21:34
add a comment |
$begingroup$
If I understand what you're asking then consider two functions:
$f: A rightarrow B, f(x)=x^2$ where $A = B = mathbb{R}$
$g: Xrightarrow Y, g(x)=x^2$ where $X = mathbb{R}$ and $Y = mathbb{R}_{geq 0}$ (real numbers $geq 0$).
Note that $-1 in A$ and $-1 in B$, but certainly there is no real number $r$ such that $r^2=-1$. As such, we say that $f$ is not onto. However, note that we still have $f(-1) = 1$, so $-1$ is an element with no pre-image but is itself the preimage of something.
Now, consider $g$ and note that $forall y in Y$, there is some $xin X$ such that $g(x)=y$, namely $x=pmsqrt{y}$. Since $g$ can map to any element of $Y$ (the co-domain), we call $g$ an "onto" or "surjective" map. Notice that simply by changing our codomain the mapping $xmapsto x^2$ can be surjective. As such, it's often very important to specify domains and codomains along with mappings when talking about functions.
Hopefully this cleared some things up!
answered Feb 1 at 22:15
Theo C.Theo C.
44228
$endgroup$
If I understand what you're asking then consider two functions:
$f: A rightarrow B, f(x)=x^2$ where $A = B = mathbb{R}$
$g: Xrightarrow Y, g(x)=x^2$ where $X = mathbb{R}$ and $Y = mathbb{R}_{geq 0}$ (real numbers $geq 0$).
Note that $-1 in A$ and $-1 in B$, but certainly there is no real number $r$ such that $r^2=-1$. As such, we say that $f$ is not onto. However, note that we still have $f(-1) = 1$, so $-1$ is an element with no pre-image but is itself the preimage of something.
Now, consider $g$ and note that $forall y in Y$, there is some $xin X$ such that $g(x)=y$, namely $x=pmsqrt{y}$. Since $g$ can map to any element of $Y$ (the co-domain), we call $g$ an "onto" or "surjective" map. Notice that simply by changing our codomain the mapping $xmapsto x^2$ can be surjective. As such, it's often very important to specify domains and codomains along with mappings when talking about functions.
Hopefully this cleared some things up!
answered Feb 1 at 22:15
Theo C.Theo C.
44228
answered Feb 1 at 22:15
Theo C.Theo C.
44228
answered Feb 1 at 22:15
Theo C.Theo C.
44228
answered Feb 1 at 22:15
Theo C.Theo C.
44228
44228
1
$begingroup$
It did thanks! I completely forgot that we are free to restrict our codomain!
$endgroup$
– houda el fezzak
Feb 3 at 21:34
add a comment |
1
$begingroup$
It did thanks! I completely forgot that we are free to restrict our codomain!
$endgroup$
– houda el fezzak
Feb 3 at 21:34
1
1
$begingroup$
It did thanks! I completely forgot that we are free to restrict our codomain!
$endgroup$
– houda el fezzak
Feb 3 at 21:34
$begingroup$
It did thanks! I completely forgot that we are free to restrict our codomain!
$endgroup$
– houda el fezzak
Feb 3 at 21:34
add a comment |
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$begingroup$
I'm a little confused about the question. You ask us to consider "some set resulting from a mapping, and that set contains some element e which has no pre-image", but how could $e$ be in such a set if it has no pre-image? Could you perhaps give a simple example?
$endgroup$
– Theo C.
Feb 1 at 21:56
$begingroup$
Oh it might be that 'resulting' is not the right expression, I rather meant.. a function which maps some elements onto another set so that there are elements in that set which have no pre-image (can't be obtained with the function?). It is absolutely possible that I am confusing many concepts and have misunderstood injective functions.
$endgroup$
– houda el fezzak
Feb 1 at 22:03