Constructing a functor that sends a monic(epic) morphism to a non-monic(epic) morphism
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The title says it all. Given two categories $C$ and $D$, I am asked to construct a funtor $F$ such that $F(f)$ is not monic (resp. epic) for a monic (resp. epic) morphism $fin C$.
My work: There is a hint that tells me to look at Monoids as a category with one element. I know that a functor from these two categories is an homomorphism from these two monoids and that the morphisms are regarded as elements of the monoid but when it comes to actually craft a functor and a morphism I have no success. I tried to consider $(mathbb{Z},.)$ as a monoid and using as functor (to itself) the trivial homomophism but given a monic morphism $f$ isnt $F(f)$ always monic? I'm getting this because $F(f)circ g_1=F(f)circ g_2=1.g_1=1.g_2$.
category-theory
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
$endgroup$
add a comment |
$begingroup$
The title says it all. Given two categories $C$ and $D$, I am asked to construct a funtor $F$ such that $F(f)$ is not monic (resp. epic) for a monic (resp. epic) morphism $fin C$.
My work: There is a hint that tells me to look at Monoids as a category with one element. I know that a functor from these two categories is an homomorphism from these two monoids and that the morphisms are regarded as elements of the monoid but when it comes to actually craft a functor and a morphism I have no success. I tried to consider $(mathbb{Z},.)$ as a monoid and using as functor (to itself) the trivial homomophism but given a monic morphism $f$ isnt $F(f)$ always monic? I'm getting this because $F(f)circ g_1=F(f)circ g_2=1.g_1=1.g_2$.
category-theory
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
$endgroup$
$begingroup$
Constant functors?
$endgroup$
– Randall
Dec 18 '18 at 2:03
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That's what I thought but either I made a mess in the second part of my attempt or this thought is wrong.
$endgroup$
– 2ndYearFreshman
Dec 18 '18 at 2:08
add a comment |
$begingroup$
The title says it all. Given two categories $C$ and $D$, I am asked to construct a funtor $F$ such that $F(f)$ is not monic (resp. epic) for a monic (resp. epic) morphism $fin C$.
My work: There is a hint that tells me to look at Monoids as a category with one element. I know that a functor from these two categories is an homomorphism from these two monoids and that the morphisms are regarded as elements of the monoid but when it comes to actually craft a functor and a morphism I have no success. I tried to consider $(mathbb{Z},.)$ as a monoid and using as functor (to itself) the trivial homomophism but given a monic morphism $f$ isnt $F(f)$ always monic? I'm getting this because $F(f)circ g_1=F(f)circ g_2=1.g_1=1.g_2$.
category-theory
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
$endgroup$
The title says it all. Given two categories $C$ and $D$, I am asked to construct a funtor $F$ such that $F(f)$ is not monic (resp. epic) for a monic (resp. epic) morphism $fin C$.
My work: There is a hint that tells me to look at Monoids as a category with one element. I know that a functor from these two categories is an homomorphism from these two monoids and that the morphisms are regarded as elements of the monoid but when it comes to actually craft a functor and a morphism I have no success. I tried to consider $(mathbb{Z},.)$ as a monoid and using as functor (to itself) the trivial homomophism but given a monic morphism $f$ isnt $F(f)$ always monic? I'm getting this because $F(f)circ g_1=F(f)circ g_2=1.g_1=1.g_2$.
category-theory
category-theory
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
asked Dec 18 '18 at 1:55
2ndYearFreshman2ndYearFreshman
125112
125112
$begingroup$
Constant functors?
$endgroup$
– Randall
Dec 18 '18 at 2:03
$begingroup$
That's what I thought but either I made a mess in the second part of my attempt or this thought is wrong.
$endgroup$
– 2ndYearFreshman
Dec 18 '18 at 2:08
add a comment |
$begingroup$
Constant functors?
$endgroup$
– Randall
Dec 18 '18 at 2:03
$begingroup$
That's what I thought but either I made a mess in the second part of my attempt or this thought is wrong.
$endgroup$
– 2ndYearFreshman
Dec 18 '18 at 2:08
$begingroup$
Constant functors?
$endgroup$
– Randall
Dec 18 '18 at 2:03
$begingroup$
Constant functors?
$endgroup$
– Randall
Dec 18 '18 at 2:03
$begingroup$
That's what I thought but either I made a mess in the second part of my attempt or this thought is wrong.
$endgroup$
– 2ndYearFreshman
Dec 18 '18 at 2:08
$begingroup$
That's what I thought but either I made a mess in the second part of my attempt or this thought is wrong.
$endgroup$
– 2ndYearFreshman
Dec 18 '18 at 2:08
add a comment |
1 Answer
1
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oldest
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I think it's much easier to find a counterexample if you look at posets, rather than monoids. Indeed in a poset every arrow is trivially a monomorphism and an epimorphism; so all you have to do is take a functor from a poset to a category, in such a way that not every arrow is sent to a monomorphism/epimorphism. For example, you can define a functor from the poset ${0leq 1}$ to the category of sets by sending $0,1$ to the set ${x,y}$, and the unique arrow $0to 1$ to the function $f$ defined by $f(x)=x=f(y)$, which is neither a mono nor an epi; this is pretty much the smallest existing counterexample.
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
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add a comment |
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1 Answer
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$begingroup$
I think it's much easier to find a counterexample if you look at posets, rather than monoids. Indeed in a poset every arrow is trivially a monomorphism and an epimorphism; so all you have to do is take a functor from a poset to a category, in such a way that not every arrow is sent to a monomorphism/epimorphism. For example, you can define a functor from the poset ${0leq 1}$ to the category of sets by sending $0,1$ to the set ${x,y}$, and the unique arrow $0to 1$ to the function $f$ defined by $f(x)=x=f(y)$, which is neither a mono nor an epi; this is pretty much the smallest existing counterexample.
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
$endgroup$
add a comment |
$begingroup$
I think it's much easier to find a counterexample if you look at posets, rather than monoids. Indeed in a poset every arrow is trivially a monomorphism and an epimorphism; so all you have to do is take a functor from a poset to a category, in such a way that not every arrow is sent to a monomorphism/epimorphism. For example, you can define a functor from the poset ${0leq 1}$ to the category of sets by sending $0,1$ to the set ${x,y}$, and the unique arrow $0to 1$ to the function $f$ defined by $f(x)=x=f(y)$, which is neither a mono nor an epi; this is pretty much the smallest existing counterexample.
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
$endgroup$
add a comment |
$begingroup$
I think it's much easier to find a counterexample if you look at posets, rather than monoids. Indeed in a poset every arrow is trivially a monomorphism and an epimorphism; so all you have to do is take a functor from a poset to a category, in such a way that not every arrow is sent to a monomorphism/epimorphism. For example, you can define a functor from the poset ${0leq 1}$ to the category of sets by sending $0,1$ to the set ${x,y}$, and the unique arrow $0to 1$ to the function $f$ defined by $f(x)=x=f(y)$, which is neither a mono nor an epi; this is pretty much the smallest existing counterexample.
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
$endgroup$
I think it's much easier to find a counterexample if you look at posets, rather than monoids. Indeed in a poset every arrow is trivially a monomorphism and an epimorphism; so all you have to do is take a functor from a poset to a category, in such a way that not every arrow is sent to a monomorphism/epimorphism. For example, you can define a functor from the poset ${0leq 1}$ to the category of sets by sending $0,1$ to the set ${x,y}$, and the unique arrow $0to 1$ to the function $f$ defined by $f(x)=x=f(y)$, which is neither a mono nor an epi; this is pretty much the smallest existing counterexample.
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
answered Dec 18 '18 at 10:18
Arnaud D.Arnaud D.
16.2k52444
16.2k52444
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$begingroup$
Constant functors?
$endgroup$
– Randall
Dec 18 '18 at 2:03
$begingroup$
That's what I thought but either I made a mess in the second part of my attempt or this thought is wrong.
$endgroup$
– 2ndYearFreshman
Dec 18 '18 at 2:08