Relation between functors under an equivalence
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Let $A$, $B$, and $C$ be three categories and $Gcolon Ato C$ and $Hcolon Bto C$ two functors. Assume that $Fcolon Ato B$ and $Ecolon Bto A$ form an equivalence between $A$ and $B$.
Suppose that
$$
H circ F = G qquad text{and} qquad G circ E = H
$$
(which should be equivalent to say that the triangle formed by $A$, $B$, and $C$ as objects and $F$, $E$, $G$, and $H$ as morphism commutes, since $F$ is an equivalence).
Is there a property, a definition, or a concept in general that describes such a relation between $H$ and $G$ under the equivalence $F$?
category-theory functors
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
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add a comment |
$begingroup$
Let $A$, $B$, and $C$ be three categories and $Gcolon Ato C$ and $Hcolon Bto C$ two functors. Assume that $Fcolon Ato B$ and $Ecolon Bto A$ form an equivalence between $A$ and $B$.
Suppose that
$$
H circ F = G qquad text{and} qquad G circ E = H
$$
(which should be equivalent to say that the triangle formed by $A$, $B$, and $C$ as objects and $F$, $E$, $G$, and $H$ as morphism commutes, since $F$ is an equivalence).
Is there a property, a definition, or a concept in general that describes such a relation between $H$ and $G$ under the equivalence $F$?
category-theory functors
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
$endgroup$
$begingroup$
What is $F^{-1}$ here? The fact that $F$ is an equivalence does not guarantee the existence of an inverse.
$endgroup$
– drhab
Dec 16 '18 at 15:55
$begingroup$
$F^{-1}$ is the pseudo-inverse (that's how it is called on the book I'm studying) of $F$ that yields the equivalence. I edit the question to make this point clear (and I change $F^{-1}$ to $E$).
$endgroup$
– BlackBrain
Dec 16 '18 at 16:06
1
$begingroup$
Are you sure you want to require equalities of functors $Hcirc F = G$ and $Gcirc E =H$ and not natural isomorphisms?
$endgroup$
– Tashi Walde
Dec 17 '18 at 18:59
add a comment |
$begingroup$
Let $A$, $B$, and $C$ be three categories and $Gcolon Ato C$ and $Hcolon Bto C$ two functors. Assume that $Fcolon Ato B$ and $Ecolon Bto A$ form an equivalence between $A$ and $B$.
Suppose that
$$
H circ F = G qquad text{and} qquad G circ E = H
$$
(which should be equivalent to say that the triangle formed by $A$, $B$, and $C$ as objects and $F$, $E$, $G$, and $H$ as morphism commutes, since $F$ is an equivalence).
Is there a property, a definition, or a concept in general that describes such a relation between $H$ and $G$ under the equivalence $F$?
category-theory functors
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
$endgroup$
Let $A$, $B$, and $C$ be three categories and $Gcolon Ato C$ and $Hcolon Bto C$ two functors. Assume that $Fcolon Ato B$ and $Ecolon Bto A$ form an equivalence between $A$ and $B$.
Suppose that
$$
H circ F = G qquad text{and} qquad G circ E = H
$$
(which should be equivalent to say that the triangle formed by $A$, $B$, and $C$ as objects and $F$, $E$, $G$, and $H$ as morphism commutes, since $F$ is an equivalence).
Is there a property, a definition, or a concept in general that describes such a relation between $H$ and $G$ under the equivalence $F$?
category-theory functors
category-theory functors
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
edited Dec 16 '18 at 16:10
BlackBrain
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
asked Dec 16 '18 at 15:50
BlackBrainBlackBrain
26416
26416
$begingroup$
What is $F^{-1}$ here? The fact that $F$ is an equivalence does not guarantee the existence of an inverse.
$endgroup$
– drhab
Dec 16 '18 at 15:55
$begingroup$
$F^{-1}$ is the pseudo-inverse (that's how it is called on the book I'm studying) of $F$ that yields the equivalence. I edit the question to make this point clear (and I change $F^{-1}$ to $E$).
$endgroup$
– BlackBrain
Dec 16 '18 at 16:06
1
$begingroup$
Are you sure you want to require equalities of functors $Hcirc F = G$ and $Gcirc E =H$ and not natural isomorphisms?
$endgroup$
– Tashi Walde
Dec 17 '18 at 18:59
add a comment |
$begingroup$
What is $F^{-1}$ here? The fact that $F$ is an equivalence does not guarantee the existence of an inverse.
$endgroup$
– drhab
Dec 16 '18 at 15:55
$begingroup$
$F^{-1}$ is the pseudo-inverse (that's how it is called on the book I'm studying) of $F$ that yields the equivalence. I edit the question to make this point clear (and I change $F^{-1}$ to $E$).
$endgroup$
– BlackBrain
Dec 16 '18 at 16:06
1
$begingroup$
Are you sure you want to require equalities of functors $Hcirc F = G$ and $Gcirc E =H$ and not natural isomorphisms?
$endgroup$
– Tashi Walde
Dec 17 '18 at 18:59
$begingroup$
What is $F^{-1}$ here? The fact that $F$ is an equivalence does not guarantee the existence of an inverse.
$endgroup$
– drhab
Dec 16 '18 at 15:55
$begingroup$
What is $F^{-1}$ here? The fact that $F$ is an equivalence does not guarantee the existence of an inverse.
$endgroup$
– drhab
Dec 16 '18 at 15:55
$begingroup$
$F^{-1}$ is the pseudo-inverse (that's how it is called on the book I'm studying) of $F$ that yields the equivalence. I edit the question to make this point clear (and I change $F^{-1}$ to $E$).
$endgroup$
– BlackBrain
Dec 16 '18 at 16:06
$begingroup$
$F^{-1}$ is the pseudo-inverse (that's how it is called on the book I'm studying) of $F$ that yields the equivalence. I edit the question to make this point clear (and I change $F^{-1}$ to $E$).
$endgroup$
– BlackBrain
Dec 16 '18 at 16:06
1
1
$begingroup$
Are you sure you want to require equalities of functors $Hcirc F = G$ and $Gcirc E =H$ and not natural isomorphisms?
$endgroup$
– Tashi Walde
Dec 17 '18 at 18:59
$begingroup$
Are you sure you want to require equalities of functors $Hcirc F = G$ and $Gcirc E =H$ and not natural isomorphisms?
$endgroup$
– Tashi Walde
Dec 17 '18 at 18:59
add a comment |
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$begingroup$
What is $F^{-1}$ here? The fact that $F$ is an equivalence does not guarantee the existence of an inverse.
$endgroup$
– drhab
Dec 16 '18 at 15:55
$begingroup$
$F^{-1}$ is the pseudo-inverse (that's how it is called on the book I'm studying) of $F$ that yields the equivalence. I edit the question to make this point clear (and I change $F^{-1}$ to $E$).
$endgroup$
– BlackBrain
Dec 16 '18 at 16:06
1
$begingroup$
Are you sure you want to require equalities of functors $Hcirc F = G$ and $Gcirc E =H$ and not natural isomorphisms?
$endgroup$
– Tashi Walde
Dec 17 '18 at 18:59