On explaining monomorphisms and epimorphisms
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I almost always see explanations of monomorphisms such as: if $forall g_1,g_2, mcirc g_1 = mcirc g_2 Rightarrow g_1=g_2$, then $m$ is monic (and similarly for epimorphisms). I find this is not at all intuitive.
Is it OK to turn the argument around and talk about inequality between arrows instead of equality? (btw, this is for a blog post, not for a scientific paper)
Monomorphisms:
$m$ is monomorphism if does not erase information: if $g_1$ and $g_2$ are different arrows, we can still tell them apart even if they are followed by $m$:
$$
g_1 neq g_2 quad Rightarrow quad m circ g_1 neq m circ g_2quad (forall g_1,g_2)
$$
In other words, the fact that $m circ g_1 = m circ g_2$ is not thanks to $m$ "throwing away" the differences between $g_1$ and $g_2$, it really must be because $g_1 = g_2$.
Epimorphisms:
$e$ is an epimorphism if it does not restrict access to information: if $g_1$ and $g_2$ are different, we can still tell them apart if they are preceded by $e$ (e.g. $e$ does not hide the parts of the source where $g_1$ and $g_2$ would have differed):
$$
g_1 neq g_2 quad Rightarrow quad g_1 circ e neq g_2 circ equad (forall g_1,g_2)
$$
In other words, the fact that $g_1circ e = g_2 circ e$ is not due to $e$ "setting things up" so that $g_1$ and $g_2$ look equal, it really must be because $g_1 = g_2$.
category-theory
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add a comment |
$begingroup$
I almost always see explanations of monomorphisms such as: if $forall g_1,g_2, mcirc g_1 = mcirc g_2 Rightarrow g_1=g_2$, then $m$ is monic (and similarly for epimorphisms). I find this is not at all intuitive.
Is it OK to turn the argument around and talk about inequality between arrows instead of equality? (btw, this is for a blog post, not for a scientific paper)
Monomorphisms:
$m$ is monomorphism if does not erase information: if $g_1$ and $g_2$ are different arrows, we can still tell them apart even if they are followed by $m$:
$$
g_1 neq g_2 quad Rightarrow quad m circ g_1 neq m circ g_2quad (forall g_1,g_2)
$$
In other words, the fact that $m circ g_1 = m circ g_2$ is not thanks to $m$ "throwing away" the differences between $g_1$ and $g_2$, it really must be because $g_1 = g_2$.
Epimorphisms:
$e$ is an epimorphism if it does not restrict access to information: if $g_1$ and $g_2$ are different, we can still tell them apart if they are preceded by $e$ (e.g. $e$ does not hide the parts of the source where $g_1$ and $g_2$ would have differed):
$$
g_1 neq g_2 quad Rightarrow quad g_1 circ e neq g_2 circ equad (forall g_1,g_2)
$$
In other words, the fact that $g_1circ e = g_2 circ e$ is not due to $e$ "setting things up" so that $g_1$ and $g_2$ look equal, it really must be because $g_1 = g_2$.
category-theory
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1
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Nothing wrong with that. Intuition might be subjective.
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– drhab
4 hours ago
add a comment |
$begingroup$
I almost always see explanations of monomorphisms such as: if $forall g_1,g_2, mcirc g_1 = mcirc g_2 Rightarrow g_1=g_2$, then $m$ is monic (and similarly for epimorphisms). I find this is not at all intuitive.
Is it OK to turn the argument around and talk about inequality between arrows instead of equality? (btw, this is for a blog post, not for a scientific paper)
Monomorphisms:
$m$ is monomorphism if does not erase information: if $g_1$ and $g_2$ are different arrows, we can still tell them apart even if they are followed by $m$:
$$
g_1 neq g_2 quad Rightarrow quad m circ g_1 neq m circ g_2quad (forall g_1,g_2)
$$
In other words, the fact that $m circ g_1 = m circ g_2$ is not thanks to $m$ "throwing away" the differences between $g_1$ and $g_2$, it really must be because $g_1 = g_2$.
Epimorphisms:
$e$ is an epimorphism if it does not restrict access to information: if $g_1$ and $g_2$ are different, we can still tell them apart if they are preceded by $e$ (e.g. $e$ does not hide the parts of the source where $g_1$ and $g_2$ would have differed):
$$
g_1 neq g_2 quad Rightarrow quad g_1 circ e neq g_2 circ equad (forall g_1,g_2)
$$
In other words, the fact that $g_1circ e = g_2 circ e$ is not due to $e$ "setting things up" so that $g_1$ and $g_2$ look equal, it really must be because $g_1 = g_2$.
category-theory
$endgroup$
I almost always see explanations of monomorphisms such as: if $forall g_1,g_2, mcirc g_1 = mcirc g_2 Rightarrow g_1=g_2$, then $m$ is monic (and similarly for epimorphisms). I find this is not at all intuitive.
Is it OK to turn the argument around and talk about inequality between arrows instead of equality? (btw, this is for a blog post, not for a scientific paper)
Monomorphisms:
$m$ is monomorphism if does not erase information: if $g_1$ and $g_2$ are different arrows, we can still tell them apart even if they are followed by $m$:
$$
g_1 neq g_2 quad Rightarrow quad m circ g_1 neq m circ g_2quad (forall g_1,g_2)
$$
In other words, the fact that $m circ g_1 = m circ g_2$ is not thanks to $m$ "throwing away" the differences between $g_1$ and $g_2$, it really must be because $g_1 = g_2$.
Epimorphisms:
$e$ is an epimorphism if it does not restrict access to information: if $g_1$ and $g_2$ are different, we can still tell them apart if they are preceded by $e$ (e.g. $e$ does not hide the parts of the source where $g_1$ and $g_2$ would have differed):
$$
g_1 neq g_2 quad Rightarrow quad g_1 circ e neq g_2 circ equad (forall g_1,g_2)
$$
In other words, the fact that $g_1circ e = g_2 circ e$ is not due to $e$ "setting things up" so that $g_1$ and $g_2$ look equal, it really must be because $g_1 = g_2$.
category-theory
category-theory
asked 4 hours ago
ZiofilZiofil
588416
588416
1
$begingroup$
Nothing wrong with that. Intuition might be subjective.
$endgroup$
– drhab
4 hours ago
add a comment |
1
$begingroup$
Nothing wrong with that. Intuition might be subjective.
$endgroup$
– drhab
4 hours ago
1
1
$begingroup$
Nothing wrong with that. Intuition might be subjective.
$endgroup$
– drhab
4 hours ago
$begingroup$
Nothing wrong with that. Intuition might be subjective.
$endgroup$
– drhab
4 hours ago
add a comment |
2 Answers
2
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$begingroup$
This is fine. Your definitions (which are the contrapositive of the conventional definitions) are logically equivalent to the conventional definitions, and if you think your definitions better supports an intuitive understanding then you are free to use them.
I would add a footnote explaining that this is not the conventional way of writing down the definition, just so your readers are prepared should they want to read up on it on their own afterwards.
$endgroup$
add a comment |
$begingroup$
This is indeed equivalent, as you just take the contrapositive of the usual definition.
I do not agree that this would be better though. We usually use the fact that an arrow is monic or epic to prove that a diagram commutes. The usual definition fits that better. However, this is personal preference and your intuition might be different. Just something you may want to take into consideration.
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add a comment |
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2 Answers
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active
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2 Answers
2
active
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votes
$begingroup$
This is fine. Your definitions (which are the contrapositive of the conventional definitions) are logically equivalent to the conventional definitions, and if you think your definitions better supports an intuitive understanding then you are free to use them.
I would add a footnote explaining that this is not the conventional way of writing down the definition, just so your readers are prepared should they want to read up on it on their own afterwards.
$endgroup$
add a comment |
$begingroup$
This is fine. Your definitions (which are the contrapositive of the conventional definitions) are logically equivalent to the conventional definitions, and if you think your definitions better supports an intuitive understanding then you are free to use them.
I would add a footnote explaining that this is not the conventional way of writing down the definition, just so your readers are prepared should they want to read up on it on their own afterwards.
$endgroup$
add a comment |
$begingroup$
This is fine. Your definitions (which are the contrapositive of the conventional definitions) are logically equivalent to the conventional definitions, and if you think your definitions better supports an intuitive understanding then you are free to use them.
I would add a footnote explaining that this is not the conventional way of writing down the definition, just so your readers are prepared should they want to read up on it on their own afterwards.
$endgroup$
This is fine. Your definitions (which are the contrapositive of the conventional definitions) are logically equivalent to the conventional definitions, and if you think your definitions better supports an intuitive understanding then you are free to use them.
I would add a footnote explaining that this is not the conventional way of writing down the definition, just so your readers are prepared should they want to read up on it on their own afterwards.
answered 4 hours ago
ArthurArthur
124k7122211
124k7122211
add a comment |
add a comment |
$begingroup$
This is indeed equivalent, as you just take the contrapositive of the usual definition.
I do not agree that this would be better though. We usually use the fact that an arrow is monic or epic to prove that a diagram commutes. The usual definition fits that better. However, this is personal preference and your intuition might be different. Just something you may want to take into consideration.
$endgroup$
add a comment |
$begingroup$
This is indeed equivalent, as you just take the contrapositive of the usual definition.
I do not agree that this would be better though. We usually use the fact that an arrow is monic or epic to prove that a diagram commutes. The usual definition fits that better. However, this is personal preference and your intuition might be different. Just something you may want to take into consideration.
$endgroup$
add a comment |
$begingroup$
This is indeed equivalent, as you just take the contrapositive of the usual definition.
I do not agree that this would be better though. We usually use the fact that an arrow is monic or epic to prove that a diagram commutes. The usual definition fits that better. However, this is personal preference and your intuition might be different. Just something you may want to take into consideration.
$endgroup$
This is indeed equivalent, as you just take the contrapositive of the usual definition.
I do not agree that this would be better though. We usually use the fact that an arrow is monic or epic to prove that a diagram commutes. The usual definition fits that better. However, this is personal preference and your intuition might be different. Just something you may want to take into consideration.
answered 4 hours ago
Mark KamsmaMark Kamsma
1,11315
1,11315
add a comment |
add a comment |
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$begingroup$
Nothing wrong with that. Intuition might be subjective.
$endgroup$
– drhab
4 hours ago