Canonical morphism from coproduct to product - Questions 1&2: Finite index sets
$begingroup$
In a pointed category $C$, for any family ${C_i}_{i in I}$ of objects, such that both their product and their coproduct exist, there is a canonical morphism
$$
varphi: coprodlimits_{i in I} C_i to prodlimits_{i in I} C_i
$$
defined by the components
$$
text{pr}_j circ varphi circ text{ins}_i = delta_{ji}
$$
for all $i,j in I$, where $delta_{ji}$ is the identity if $i = j$ and the zero morphism otherwise. If this morphism is an isomorphism for finite $I$, the (co) product is called a biproduct, but apart from this important case I find very little information about it in the literature.
By many examples, I see the following for finite $I$:
$varphi$ can be monic (e.g. $text{Set}_*, text{Mod}_R$) or not ($text{Group}$).
$varphi$ can be epic (algebraic categories like $text{Group}, text{Mod}_R$ etc.) or not (e.g. $text{Set}_*$).
$varphi$ can even be an isomorphism (e.g. $text{Mod}_R$).
Among the possible combinations, there are two that I could not find. These constitute my questions:
- Is there a case where $varphi$ is neither epic nor monic?
- Is there a case where $varphi$ is a non-isic bimorphism?
category-theory
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
$endgroup$
add a comment |
$begingroup$
In a pointed category $C$, for any family ${C_i}_{i in I}$ of objects, such that both their product and their coproduct exist, there is a canonical morphism
$$
varphi: coprodlimits_{i in I} C_i to prodlimits_{i in I} C_i
$$
defined by the components
$$
text{pr}_j circ varphi circ text{ins}_i = delta_{ji}
$$
for all $i,j in I$, where $delta_{ji}$ is the identity if $i = j$ and the zero morphism otherwise. If this morphism is an isomorphism for finite $I$, the (co) product is called a biproduct, but apart from this important case I find very little information about it in the literature.
By many examples, I see the following for finite $I$:
$varphi$ can be monic (e.g. $text{Set}_*, text{Mod}_R$) or not ($text{Group}$).
$varphi$ can be epic (algebraic categories like $text{Group}, text{Mod}_R$ etc.) or not (e.g. $text{Set}_*$).
$varphi$ can even be an isomorphism (e.g. $text{Mod}_R$).
Among the possible combinations, there are two that I could not find. These constitute my questions:
- Is there a case where $varphi$ is neither epic nor monic?
- Is there a case where $varphi$ is a non-isic bimorphism?
category-theory
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
$endgroup$
add a comment |
$begingroup$
In a pointed category $C$, for any family ${C_i}_{i in I}$ of objects, such that both their product and their coproduct exist, there is a canonical morphism
$$
varphi: coprodlimits_{i in I} C_i to prodlimits_{i in I} C_i
$$
defined by the components
$$
text{pr}_j circ varphi circ text{ins}_i = delta_{ji}
$$
for all $i,j in I$, where $delta_{ji}$ is the identity if $i = j$ and the zero morphism otherwise. If this morphism is an isomorphism for finite $I$, the (co) product is called a biproduct, but apart from this important case I find very little information about it in the literature.
By many examples, I see the following for finite $I$:
$varphi$ can be monic (e.g. $text{Set}_*, text{Mod}_R$) or not ($text{Group}$).
$varphi$ can be epic (algebraic categories like $text{Group}, text{Mod}_R$ etc.) or not (e.g. $text{Set}_*$).
$varphi$ can even be an isomorphism (e.g. $text{Mod}_R$).
Among the possible combinations, there are two that I could not find. These constitute my questions:
- Is there a case where $varphi$ is neither epic nor monic?
- Is there a case where $varphi$ is a non-isic bimorphism?
category-theory
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
$endgroup$
In a pointed category $C$, for any family ${C_i}_{i in I}$ of objects, such that both their product and their coproduct exist, there is a canonical morphism
$$
varphi: coprodlimits_{i in I} C_i to prodlimits_{i in I} C_i
$$
defined by the components
$$
text{pr}_j circ varphi circ text{ins}_i = delta_{ji}
$$
for all $i,j in I$, where $delta_{ji}$ is the identity if $i = j$ and the zero morphism otherwise. If this morphism is an isomorphism for finite $I$, the (co) product is called a biproduct, but apart from this important case I find very little information about it in the literature.
By many examples, I see the following for finite $I$:
$varphi$ can be monic (e.g. $text{Set}_*, text{Mod}_R$) or not ($text{Group}$).
$varphi$ can be epic (algebraic categories like $text{Group}, text{Mod}_R$ etc.) or not (e.g. $text{Set}_*$).
$varphi$ can even be an isomorphism (e.g. $text{Mod}_R$).
Among the possible combinations, there are two that I could not find. These constitute my questions:
- Is there a case where $varphi$ is neither epic nor monic?
- Is there a case where $varphi$ is a non-isic bimorphism?
category-theory
category-theory
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
edited Nov 30 '18 at 16:29
Gnampfissimo
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
asked Nov 30 '18 at 16:00
GnampfissimoGnampfissimo
18011
18011
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, there's sort of a lame way to do this by taking the product of two of your existing examples, e.g. $text{Group} times text{Set}_{ast}$.
Yes. Consider the category $text{Ban}_1$ of Banach spaces and weak contractions (morphisms of norm $le 1$). The finite coproduct and product in this category are both the direct sum, but with different norms: the finite coproduct has an "$ell^1$ norm" and the finite product has an "$ell^{infty}$ norm." The underlying map of sets from the coproduct to the product is a bijection, so this map is a bimorphism, but it's not an isomorphism.
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
$endgroup$
add a comment |
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$begingroup$
Yes, there's sort of a lame way to do this by taking the product of two of your existing examples, e.g. $text{Group} times text{Set}_{ast}$.
Yes. Consider the category $text{Ban}_1$ of Banach spaces and weak contractions (morphisms of norm $le 1$). The finite coproduct and product in this category are both the direct sum, but with different norms: the finite coproduct has an "$ell^1$ norm" and the finite product has an "$ell^{infty}$ norm." The underlying map of sets from the coproduct to the product is a bijection, so this map is a bimorphism, but it's not an isomorphism.
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
$endgroup$
add a comment |
$begingroup$
Yes, there's sort of a lame way to do this by taking the product of two of your existing examples, e.g. $text{Group} times text{Set}_{ast}$.
Yes. Consider the category $text{Ban}_1$ of Banach spaces and weak contractions (morphisms of norm $le 1$). The finite coproduct and product in this category are both the direct sum, but with different norms: the finite coproduct has an "$ell^1$ norm" and the finite product has an "$ell^{infty}$ norm." The underlying map of sets from the coproduct to the product is a bijection, so this map is a bimorphism, but it's not an isomorphism.
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
$endgroup$
add a comment |
$begingroup$
Yes, there's sort of a lame way to do this by taking the product of two of your existing examples, e.g. $text{Group} times text{Set}_{ast}$.
Yes. Consider the category $text{Ban}_1$ of Banach spaces and weak contractions (morphisms of norm $le 1$). The finite coproduct and product in this category are both the direct sum, but with different norms: the finite coproduct has an "$ell^1$ norm" and the finite product has an "$ell^{infty}$ norm." The underlying map of sets from the coproduct to the product is a bijection, so this map is a bimorphism, but it's not an isomorphism.
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
$endgroup$
Yes, there's sort of a lame way to do this by taking the product of two of your existing examples, e.g. $text{Group} times text{Set}_{ast}$.
Yes. Consider the category $text{Ban}_1$ of Banach spaces and weak contractions (morphisms of norm $le 1$). The finite coproduct and product in this category are both the direct sum, but with different norms: the finite coproduct has an "$ell^1$ norm" and the finite product has an "$ell^{infty}$ norm." The underlying map of sets from the coproduct to the product is a bijection, so this map is a bimorphism, but it's not an isomorphism.
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
edited Nov 30 '18 at 19:55
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
answered Nov 30 '18 at 19:21
Qiaochu YuanQiaochu Yuan
278k32584919
278k32584919
add a comment |
add a comment |
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