Having trouble proving natural transformation horizontal composition equality of two formulas using a...
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Let $C, C', C''$ be three categories. Let $C xrightarrow{F_1, F_2} C'$ and $C' xrightarrow{G_1, G_2} C''$, be four functors, and let $theta : F_1 Rightarrow F_2, lambda : G_1 Rightarrow G_2$, be two natural transformations. Then I've already shown algebraically that if we defined $(lambda * theta)_X equiv G_2(theta_X) circ lambda_{F_1 X}$, then $lambda * theta$ is a natural transformation $G_1 circ F_1 Rightarrow G_2 circ F_2$. Now I want to show that for all $X in C$, the following diagram commutes (or the two formulas for $lambda * theta$ are the same. A book says by naturality of $theta$ and functoriality of $G_1$ the square commutes. But I'm having trouble seeing this. Please explain using diagrams or give a hint.
abstract-algebra category-theory proof-explanation
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
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add a comment |
$begingroup$
Let $C, C', C''$ be three categories. Let $C xrightarrow{F_1, F_2} C'$ and $C' xrightarrow{G_1, G_2} C''$, be four functors, and let $theta : F_1 Rightarrow F_2, lambda : G_1 Rightarrow G_2$, be two natural transformations. Then I've already shown algebraically that if we defined $(lambda * theta)_X equiv G_2(theta_X) circ lambda_{F_1 X}$, then $lambda * theta$ is a natural transformation $G_1 circ F_1 Rightarrow G_2 circ F_2$. Now I want to show that for all $X in C$, the following diagram commutes (or the two formulas for $lambda * theta$ are the same. A book says by naturality of $theta$ and functoriality of $G_1$ the square commutes. But I'm having trouble seeing this. Please explain using diagrams or give a hint.
abstract-algebra category-theory proof-explanation
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
$begingroup$
No rush on this. Got to do some work any way!
$endgroup$
– Hermit with Adjoint
Mar 16 '16 at 18:15
add a comment |
$begingroup$
Let $C, C', C''$ be three categories. Let $C xrightarrow{F_1, F_2} C'$ and $C' xrightarrow{G_1, G_2} C''$, be four functors, and let $theta : F_1 Rightarrow F_2, lambda : G_1 Rightarrow G_2$, be two natural transformations. Then I've already shown algebraically that if we defined $(lambda * theta)_X equiv G_2(theta_X) circ lambda_{F_1 X}$, then $lambda * theta$ is a natural transformation $G_1 circ F_1 Rightarrow G_2 circ F_2$. Now I want to show that for all $X in C$, the following diagram commutes (or the two formulas for $lambda * theta$ are the same. A book says by naturality of $theta$ and functoriality of $G_1$ the square commutes. But I'm having trouble seeing this. Please explain using diagrams or give a hint.
abstract-algebra category-theory proof-explanation
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
Let $C, C', C''$ be three categories. Let $C xrightarrow{F_1, F_2} C'$ and $C' xrightarrow{G_1, G_2} C''$, be four functors, and let $theta : F_1 Rightarrow F_2, lambda : G_1 Rightarrow G_2$, be two natural transformations. Then I've already shown algebraically that if we defined $(lambda * theta)_X equiv G_2(theta_X) circ lambda_{F_1 X}$, then $lambda * theta$ is a natural transformation $G_1 circ F_1 Rightarrow G_2 circ F_2$. Now I want to show that for all $X in C$, the following diagram commutes (or the two formulas for $lambda * theta$ are the same. A book says by naturality of $theta$ and functoriality of $G_1$ the square commutes. But I'm having trouble seeing this. Please explain using diagrams or give a hint.
abstract-algebra category-theory proof-explanation
abstract-algebra category-theory proof-explanation
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
asked Mar 16 '16 at 18:03
Hermit with AdjointHermit with Adjoint
9,11552458
9,11552458
$begingroup$
No rush on this. Got to do some work any way!
$endgroup$
– Hermit with Adjoint
Mar 16 '16 at 18:15
add a comment |
$begingroup$
No rush on this. Got to do some work any way!
$endgroup$
– Hermit with Adjoint
Mar 16 '16 at 18:15
$begingroup$
No rush on this. Got to do some work any way!
$endgroup$
– Hermit with Adjoint
Mar 16 '16 at 18:15
$begingroup$
No rush on this. Got to do some work any way!
$endgroup$
– Hermit with Adjoint
Mar 16 '16 at 18:15
add a comment |
2 Answers
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You have already written the proof of this fact.
Definition. Let $mathcal{A}$ and $mathcal{B}$ be categories, $T,Scolonmathcal{A}tomathcal{B}$ be functors, $alphacolontext{Obj}(mathcal{A})totext{Mor}(mathcal{B})$ be a function. Then the function $alpha$ is called a natural transformation iff for every morphism $fcolon a_1to a_2$ in $mathcal{A}$ the following equality holds:
$$
S(f)circalpha(a_1)=alpha(a_2)circ T(f).
$$
Now assign $mathcal{A}=C'$, $mathcal{B}=C''$, $T=G_1$, $S=G_2$, $alpha=lambda$, $a_1=F_1(x)$, $a_2=F_2(x)$, $f=theta(x)$ and we get:
$$
G_2(theta(x))circlambda(F_1(X))=lambda(F_2(x))circ G_1(theta(x)),
$$
as required. The commutativity of your diagram is nothing but this equality.
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
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add a comment |
$begingroup$
Here's a more diagramatically-based proof.
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
You have already written the proof of this fact.
Definition. Let $mathcal{A}$ and $mathcal{B}$ be categories, $T,Scolonmathcal{A}tomathcal{B}$ be functors, $alphacolontext{Obj}(mathcal{A})totext{Mor}(mathcal{B})$ be a function. Then the function $alpha$ is called a natural transformation iff for every morphism $fcolon a_1to a_2$ in $mathcal{A}$ the following equality holds:
$$
S(f)circalpha(a_1)=alpha(a_2)circ T(f).
$$
Now assign $mathcal{A}=C'$, $mathcal{B}=C''$, $T=G_1$, $S=G_2$, $alpha=lambda$, $a_1=F_1(x)$, $a_2=F_2(x)$, $f=theta(x)$ and we get:
$$
G_2(theta(x))circlambda(F_1(X))=lambda(F_2(x))circ G_1(theta(x)),
$$
as required. The commutativity of your diagram is nothing but this equality.
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
$endgroup$
add a comment |
$begingroup$
You have already written the proof of this fact.
Definition. Let $mathcal{A}$ and $mathcal{B}$ be categories, $T,Scolonmathcal{A}tomathcal{B}$ be functors, $alphacolontext{Obj}(mathcal{A})totext{Mor}(mathcal{B})$ be a function. Then the function $alpha$ is called a natural transformation iff for every morphism $fcolon a_1to a_2$ in $mathcal{A}$ the following equality holds:
$$
S(f)circalpha(a_1)=alpha(a_2)circ T(f).
$$
Now assign $mathcal{A}=C'$, $mathcal{B}=C''$, $T=G_1$, $S=G_2$, $alpha=lambda$, $a_1=F_1(x)$, $a_2=F_2(x)$, $f=theta(x)$ and we get:
$$
G_2(theta(x))circlambda(F_1(X))=lambda(F_2(x))circ G_1(theta(x)),
$$
as required. The commutativity of your diagram is nothing but this equality.
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
$endgroup$
add a comment |
$begingroup$
You have already written the proof of this fact.
Definition. Let $mathcal{A}$ and $mathcal{B}$ be categories, $T,Scolonmathcal{A}tomathcal{B}$ be functors, $alphacolontext{Obj}(mathcal{A})totext{Mor}(mathcal{B})$ be a function. Then the function $alpha$ is called a natural transformation iff for every morphism $fcolon a_1to a_2$ in $mathcal{A}$ the following equality holds:
$$
S(f)circalpha(a_1)=alpha(a_2)circ T(f).
$$
Now assign $mathcal{A}=C'$, $mathcal{B}=C''$, $T=G_1$, $S=G_2$, $alpha=lambda$, $a_1=F_1(x)$, $a_2=F_2(x)$, $f=theta(x)$ and we get:
$$
G_2(theta(x))circlambda(F_1(X))=lambda(F_2(x))circ G_1(theta(x)),
$$
as required. The commutativity of your diagram is nothing but this equality.
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
$endgroup$
You have already written the proof of this fact.
Definition. Let $mathcal{A}$ and $mathcal{B}$ be categories, $T,Scolonmathcal{A}tomathcal{B}$ be functors, $alphacolontext{Obj}(mathcal{A})totext{Mor}(mathcal{B})$ be a function. Then the function $alpha$ is called a natural transformation iff for every morphism $fcolon a_1to a_2$ in $mathcal{A}$ the following equality holds:
$$
S(f)circalpha(a_1)=alpha(a_2)circ T(f).
$$
Now assign $mathcal{A}=C'$, $mathcal{B}=C''$, $T=G_1$, $S=G_2$, $alpha=lambda$, $a_1=F_1(x)$, $a_2=F_2(x)$, $f=theta(x)$ and we get:
$$
G_2(theta(x))circlambda(F_1(X))=lambda(F_2(x))circ G_1(theta(x)),
$$
as required. The commutativity of your diagram is nothing but this equality.
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
edited Mar 16 '16 at 20:14
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
answered Mar 16 '16 at 18:19
OskarOskar
3,0781719
3,0781719
add a comment |
add a comment |
$begingroup$
Here's a more diagramatically-based proof.
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
add a comment |
$begingroup$
Here's a more diagramatically-based proof.
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
add a comment |
$begingroup$
Here's a more diagramatically-based proof.
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
Here's a more diagramatically-based proof.
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
answered Mar 17 '16 at 18:13
Hermit with AdjointHermit with Adjoint
9,11552458
9,11552458
add a comment |
add a comment |
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$begingroup$
No rush on this. Got to do some work any way!
$endgroup$
– Hermit with Adjoint
Mar 16 '16 at 18:15