Mixing
Proving equivalent categories have binary products if one of them does
Suppose $mathscr{C},mathscr{D}$ are equivalent categories.
Then there exist functors $S∶ mathscr{C}rightarrow mathscr{D}$ and $T∶mathscr{D}→mathscr{C}$ with the compositions defined $Tcirc S∶ mathscr{C}→mathscr{C}$ and $Scirc T∶ mathscr{D}rightarrowmathscr{D}$, and a pair of natural isomorphisms $alpha∶Tcirc S rightarrow 1_mathscr{C}$ and $beta ∶Scirc Trightarrow1_mathscr{D}.$
We seek to prove that, if $mathscr{C} $ has binary products, then $mathscr{D} $ also does.
I'm not really sure how to even approach this one, there seems to be so much going on that it's almost dizzying. My gut instinct is that we could show $mathscr{C}$ has limits (since aren't limits a way of formalizing the product?) and then ... somehow? ... show that $mathscr{D}$ also has them, but I'm so unconfident in that even conceptually.
Any ideas or nudges?
category-theory natural-transformations
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
|
show 4 more comments
Suppose $mathscr{C},mathscr{D}$ are equivalent categories.
Then there exist functors $S∶ mathscr{C}rightarrow mathscr{D}$ and $T∶mathscr{D}→mathscr{C}$ with the compositions defined $Tcirc S∶ mathscr{C}→mathscr{C}$ and $Scirc T∶ mathscr{D}rightarrowmathscr{D}$, and a pair of natural isomorphisms $alpha∶Tcirc S rightarrow 1_mathscr{C}$ and $beta ∶Scirc Trightarrow1_mathscr{D}.$
We seek to prove that, if $mathscr{C} $ has binary products, then $mathscr{D} $ also does.
I'm not really sure how to even approach this one, there seems to be so much going on that it's almost dizzying. My gut instinct is that we could show $mathscr{C}$ has limits (since aren't limits a way of formalizing the product?) and then ... somehow? ... show that $mathscr{D}$ also has them, but I'm so unconfident in that even conceptually.
Any ideas or nudges?
category-theory natural-transformations
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
No, you can't just magically make arbitrary limits out of binary products.
– darij grinberg
Nov 20 '18 at 5:52
Okay then; what am I supposed to do in this?
– Eevee Trainer
Nov 20 '18 at 6:23
So you want to construct a product of two objects $X$ and $Y$ of $mathscr{D}$. You can construct binary products in $mathscr{C}$, so you can construct a product of $Tleft(Xright)$ and $Tleft(Yright)$. Furthermore, you can send this product back into $mathscr{D}$ using $S$. Does this do the trick?
– darij grinberg
Nov 20 '18 at 6:24
So $X,Y$ have image $T(X), T(Y)$ in $mathscr{C}$. Since $mathscr{C}$, by assumption, has binary products, then the object $T(X) times T(Y)$ must also exist in $mathscr{C}$. ... I'm a bit lost from there though. Obviously we can just send that product through $S$ back to $mathscr{D}$, but that doesn't really invoke any assumptions about the equivalence of the categories, namely the natural transformations (among other probable issues). So it's obviously more nuanced than that.
– Eevee Trainer
Nov 20 '18 at 6:48
@EeveeTrainer No, that's the right idea. The natural transformations go to show that $S$ sends the product to a product of $STX$ and $STY$, and thus of $X$ and $Y$.
– Kevin Carlson
Nov 20 '18 at 17:25
|
show 4 more comments
Suppose $mathscr{C},mathscr{D}$ are equivalent categories.
Then there exist functors $S∶ mathscr{C}rightarrow mathscr{D}$ and $T∶mathscr{D}→mathscr{C}$ with the compositions defined $Tcirc S∶ mathscr{C}→mathscr{C}$ and $Scirc T∶ mathscr{D}rightarrowmathscr{D}$, and a pair of natural isomorphisms $alpha∶Tcirc S rightarrow 1_mathscr{C}$ and $beta ∶Scirc Trightarrow1_mathscr{D}.$
We seek to prove that, if $mathscr{C} $ has binary products, then $mathscr{D} $ also does.
I'm not really sure how to even approach this one, there seems to be so much going on that it's almost dizzying. My gut instinct is that we could show $mathscr{C}$ has limits (since aren't limits a way of formalizing the product?) and then ... somehow? ... show that $mathscr{D}$ also has them, but I'm so unconfident in that even conceptually.
Any ideas or nudges?
category-theory natural-transformations
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
Suppose $mathscr{C},mathscr{D}$ are equivalent categories.
Then there exist functors $S∶ mathscr{C}rightarrow mathscr{D}$ and $T∶mathscr{D}→mathscr{C}$ with the compositions defined $Tcirc S∶ mathscr{C}→mathscr{C}$ and $Scirc T∶ mathscr{D}rightarrowmathscr{D}$, and a pair of natural isomorphisms $alpha∶Tcirc S rightarrow 1_mathscr{C}$ and $beta ∶Scirc Trightarrow1_mathscr{D}.$
We seek to prove that, if $mathscr{C} $ has binary products, then $mathscr{D} $ also does.
I'm not really sure how to even approach this one, there seems to be so much going on that it's almost dizzying. My gut instinct is that we could show $mathscr{C}$ has limits (since aren't limits a way of formalizing the product?) and then ... somehow? ... show that $mathscr{D}$ also has them, but I'm so unconfident in that even conceptually.
Any ideas or nudges?
category-theory natural-transformations
category-theory natural-transformations
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
edited Nov 20 '18 at 15:30
darij grinberg
10.2k33062
10.2k33062
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
asked Nov 20 '18 at 5:44
Eevee Trainer
4,6121634
4,6121634
No, you can't just magically make arbitrary limits out of binary products.
– darij grinberg
Nov 20 '18 at 5:52
Okay then; what am I supposed to do in this?
– Eevee Trainer
Nov 20 '18 at 6:23
So you want to construct a product of two objects $X$ and $Y$ of $mathscr{D}$. You can construct binary products in $mathscr{C}$, so you can construct a product of $Tleft(Xright)$ and $Tleft(Yright)$. Furthermore, you can send this product back into $mathscr{D}$ using $S$. Does this do the trick?
– darij grinberg
Nov 20 '18 at 6:24
So $X,Y$ have image $T(X), T(Y)$ in $mathscr{C}$. Since $mathscr{C}$, by assumption, has binary products, then the object $T(X) times T(Y)$ must also exist in $mathscr{C}$. ... I'm a bit lost from there though. Obviously we can just send that product through $S$ back to $mathscr{D}$, but that doesn't really invoke any assumptions about the equivalence of the categories, namely the natural transformations (among other probable issues). So it's obviously more nuanced than that.
– Eevee Trainer
Nov 20 '18 at 6:48
@EeveeTrainer No, that's the right idea. The natural transformations go to show that $S$ sends the product to a product of $STX$ and $STY$, and thus of $X$ and $Y$.
– Kevin Carlson
Nov 20 '18 at 17:25
|
show 4 more comments
No, you can't just magically make arbitrary limits out of binary products.
– darij grinberg
Nov 20 '18 at 5:52
Okay then; what am I supposed to do in this?
– Eevee Trainer
Nov 20 '18 at 6:23
So you want to construct a product of two objects $X$ and $Y$ of $mathscr{D}$. You can construct binary products in $mathscr{C}$, so you can construct a product of $Tleft(Xright)$ and $Tleft(Yright)$. Furthermore, you can send this product back into $mathscr{D}$ using $S$. Does this do the trick?
– darij grinberg
Nov 20 '18 at 6:24
So $X,Y$ have image $T(X), T(Y)$ in $mathscr{C}$. Since $mathscr{C}$, by assumption, has binary products, then the object $T(X) times T(Y)$ must also exist in $mathscr{C}$. ... I'm a bit lost from there though. Obviously we can just send that product through $S$ back to $mathscr{D}$, but that doesn't really invoke any assumptions about the equivalence of the categories, namely the natural transformations (among other probable issues). So it's obviously more nuanced than that.
– Eevee Trainer
Nov 20 '18 at 6:48
@EeveeTrainer No, that's the right idea. The natural transformations go to show that $S$ sends the product to a product of $STX$ and $STY$, and thus of $X$ and $Y$.
– Kevin Carlson
Nov 20 '18 at 17:25
No, you can't just magically make arbitrary limits out of binary products.
– darij grinberg
Nov 20 '18 at 5:52
No, you can't just magically make arbitrary limits out of binary products.
– darij grinberg
Nov 20 '18 at 5:52
Okay then; what am I supposed to do in this?
– Eevee Trainer
Nov 20 '18 at 6:23
Okay then; what am I supposed to do in this?
– Eevee Trainer
Nov 20 '18 at 6:23
So you want to construct a product of two objects $X$ and $Y$ of $mathscr{D}$. You can construct binary products in $mathscr{C}$, so you can construct a product of $Tleft(Xright)$ and $Tleft(Yright)$. Furthermore, you can send this product back into $mathscr{D}$ using $S$. Does this do the trick?
– darij grinberg
Nov 20 '18 at 6:24
So you want to construct a product of two objects $X$ and $Y$ of $mathscr{D}$. You can construct binary products in $mathscr{C}$, so you can construct a product of $Tleft(Xright)$ and $Tleft(Yright)$. Furthermore, you can send this product back into $mathscr{D}$ using $S$. Does this do the trick?
– darij grinberg
Nov 20 '18 at 6:24
So $X,Y$ have image $T(X), T(Y)$ in $mathscr{C}$. Since $mathscr{C}$, by assumption, has binary products, then the object $T(X) times T(Y)$ must also exist in $mathscr{C}$. ... I'm a bit lost from there though. Obviously we can just send that product through $S$ back to $mathscr{D}$, but that doesn't really invoke any assumptions about the equivalence of the categories, namely the natural transformations (among other probable issues). So it's obviously more nuanced than that.
– Eevee Trainer
Nov 20 '18 at 6:48
So $X,Y$ have image $T(X), T(Y)$ in $mathscr{C}$. Since $mathscr{C}$, by assumption, has binary products, then the object $T(X) times T(Y)$ must also exist in $mathscr{C}$. ... I'm a bit lost from there though. Obviously we can just send that product through $S$ back to $mathscr{D}$, but that doesn't really invoke any assumptions about the equivalence of the categories, namely the natural transformations (among other probable issues). So it's obviously more nuanced than that.
– Eevee Trainer
Nov 20 '18 at 6:48
@EeveeTrainer No, that's the right idea. The natural transformations go to show that $S$ sends the product to a product of $STX$ and $STY$, and thus of $X$ and $Y$.
– Kevin Carlson
Nov 20 '18 at 17:25
@EeveeTrainer No, that's the right idea. The natural transformations go to show that $S$ sends the product to a product of $STX$ and $STY$, and thus of $X$ and $Y$.
– Kevin Carlson
Nov 20 '18 at 17:25
|
show 4 more comments
1 Answer
1
active
oldest
votes
Let $X,Y$ be objects in $mathscr D$ and
begin{align}
&p:T(X)times T(Y)to T(X)&
&q:T(X)times T(Y)to T(Y)
end{align}
be a product source in $mathscr C$.
We claim that
begin{align}
&S(p)beta_X:S(T(X)times T(Y))to X&
&S(q)beta_Y:S(T(X)times T(Y))to Y
end{align}
is a product source in $mathscr D$.
Let
begin{align}
&u:Zto X&
&v:Zto Y
end{align}
be a source in $mathscr D$.
Then
begin{align}
&T(u):T(Z)to T(X)&
&T(v):T(Z)to T(Y)
end{align}
is a source in $mathscr C$, hence there exists one and only one morphism $w:T(Z)to T(X)times T(Y)$ such that
begin{align}
&T(u)=wp&
&T(v)=wq
end{align}
Then
begin{align}
beta^{-1}_ZS(w)S(p)beta_X&=beta^{-1}_Z(Scirc T)(u)beta_X&
beta^{-1}_ZS(w)S(q)beta_Y&=beta^{-1}_Z(Scirc T)(v)beta_Y\
&=beta^{-1}_Zbeta_Zu&
&=beta^{-1}_Zbeta_Zv\
&=u&
&=v
end{align}
which proves the existence part.
It remains to prove uniqueness.
For this, first recall that we can assume $alpha^{-1}_TT(beta)=1_T$ and $beta^{-1}_SS(alpha)=1_S$ (see, for example, here).
Let $w':Zto S(T(X)times T(Y))$ such that
begin{align}
&u=w'S(p)beta_X&
&v=w'S(q)beta_Y
end{align}
Then
begin{align}
T(u)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}p&
T(v)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}q\
&=T(w')alpha_{T(X)times T(Y)}p&
&=T(w')alpha_{T(X)times T(Y)}q\
end{align}
from which $w=T(w')alpha_{T(X)times T(Y)}$.
Consequently,
begin{align}
S(w)
&=beta_Zbeta^{-1}_Z(Scirc T)(w')S(alpha_{T(X)times T(Y)})\
&=beta_Zw'beta^{-1}_{S(T(X)times T(Y))}S(alpha_{T(X)times T(Y)})\
&=beta_Zw'
end{align}
thus concluding the proof.
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
Hmm. Your "source" means "cone in the particular case of a binary product", right? And you're writing composition the "other way round"?
– darij grinberg
Nov 20 '18 at 15:26
Yes, you are right;
– Fabio Lucchini
Nov 20 '18 at 16:13
@FabioLucchini I think beginning students should be warned about a choice like reversing the traditional composition order!
– Kevin Carlson
Nov 20 '18 at 17:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $X,Y$ be objects in $mathscr D$ and
begin{align}
&p:T(X)times T(Y)to T(X)&
&q:T(X)times T(Y)to T(Y)
end{align}
be a product source in $mathscr C$.
We claim that
begin{align}
&S(p)beta_X:S(T(X)times T(Y))to X&
&S(q)beta_Y:S(T(X)times T(Y))to Y
end{align}
is a product source in $mathscr D$.
Let
begin{align}
&u:Zto X&
&v:Zto Y
end{align}
be a source in $mathscr D$.
Then
begin{align}
&T(u):T(Z)to T(X)&
&T(v):T(Z)to T(Y)
end{align}
is a source in $mathscr C$, hence there exists one and only one morphism $w:T(Z)to T(X)times T(Y)$ such that
begin{align}
&T(u)=wp&
&T(v)=wq
end{align}
Then
begin{align}
beta^{-1}_ZS(w)S(p)beta_X&=beta^{-1}_Z(Scirc T)(u)beta_X&
beta^{-1}_ZS(w)S(q)beta_Y&=beta^{-1}_Z(Scirc T)(v)beta_Y\
&=beta^{-1}_Zbeta_Zu&
&=beta^{-1}_Zbeta_Zv\
&=u&
&=v
end{align}
which proves the existence part.
It remains to prove uniqueness.
For this, first recall that we can assume $alpha^{-1}_TT(beta)=1_T$ and $beta^{-1}_SS(alpha)=1_S$ (see, for example, here).
Let $w':Zto S(T(X)times T(Y))$ such that
begin{align}
&u=w'S(p)beta_X&
&v=w'S(q)beta_Y
end{align}
Then
begin{align}
T(u)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}p&
T(v)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}q\
&=T(w')alpha_{T(X)times T(Y)}p&
&=T(w')alpha_{T(X)times T(Y)}q\
end{align}
from which $w=T(w')alpha_{T(X)times T(Y)}$.
Consequently,
begin{align}
S(w)
&=beta_Zbeta^{-1}_Z(Scirc T)(w')S(alpha_{T(X)times T(Y)})\
&=beta_Zw'beta^{-1}_{S(T(X)times T(Y))}S(alpha_{T(X)times T(Y)})\
&=beta_Zw'
end{align}
thus concluding the proof.
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
Hmm. Your "source" means "cone in the particular case of a binary product", right? And you're writing composition the "other way round"?
– darij grinberg
Nov 20 '18 at 15:26
Yes, you are right;
– Fabio Lucchini
Nov 20 '18 at 16:13
@FabioLucchini I think beginning students should be warned about a choice like reversing the traditional composition order!
– Kevin Carlson
Nov 20 '18 at 17:26
add a comment |
Let $X,Y$ be objects in $mathscr D$ and
begin{align}
&p:T(X)times T(Y)to T(X)&
&q:T(X)times T(Y)to T(Y)
end{align}
be a product source in $mathscr C$.
We claim that
begin{align}
&S(p)beta_X:S(T(X)times T(Y))to X&
&S(q)beta_Y:S(T(X)times T(Y))to Y
end{align}
is a product source in $mathscr D$.
Let
begin{align}
&u:Zto X&
&v:Zto Y
end{align}
be a source in $mathscr D$.
Then
begin{align}
&T(u):T(Z)to T(X)&
&T(v):T(Z)to T(Y)
end{align}
is a source in $mathscr C$, hence there exists one and only one morphism $w:T(Z)to T(X)times T(Y)$ such that
begin{align}
&T(u)=wp&
&T(v)=wq
end{align}
Then
begin{align}
beta^{-1}_ZS(w)S(p)beta_X&=beta^{-1}_Z(Scirc T)(u)beta_X&
beta^{-1}_ZS(w)S(q)beta_Y&=beta^{-1}_Z(Scirc T)(v)beta_Y\
&=beta^{-1}_Zbeta_Zu&
&=beta^{-1}_Zbeta_Zv\
&=u&
&=v
end{align}
which proves the existence part.
It remains to prove uniqueness.
For this, first recall that we can assume $alpha^{-1}_TT(beta)=1_T$ and $beta^{-1}_SS(alpha)=1_S$ (see, for example, here).
Let $w':Zto S(T(X)times T(Y))$ such that
begin{align}
&u=w'S(p)beta_X&
&v=w'S(q)beta_Y
end{align}
Then
begin{align}
T(u)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}p&
T(v)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}q\
&=T(w')alpha_{T(X)times T(Y)}p&
&=T(w')alpha_{T(X)times T(Y)}q\
end{align}
from which $w=T(w')alpha_{T(X)times T(Y)}$.
Consequently,
begin{align}
S(w)
&=beta_Zbeta^{-1}_Z(Scirc T)(w')S(alpha_{T(X)times T(Y)})\
&=beta_Zw'beta^{-1}_{S(T(X)times T(Y))}S(alpha_{T(X)times T(Y)})\
&=beta_Zw'
end{align}
thus concluding the proof.
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
Hmm. Your "source" means "cone in the particular case of a binary product", right? And you're writing composition the "other way round"?
– darij grinberg
Nov 20 '18 at 15:26
Yes, you are right;
– Fabio Lucchini
Nov 20 '18 at 16:13
@FabioLucchini I think beginning students should be warned about a choice like reversing the traditional composition order!
– Kevin Carlson
Nov 20 '18 at 17:26
add a comment |
Let $X,Y$ be objects in $mathscr D$ and
begin{align}
&p:T(X)times T(Y)to T(X)&
&q:T(X)times T(Y)to T(Y)
end{align}
be a product source in $mathscr C$.
We claim that
begin{align}
&S(p)beta_X:S(T(X)times T(Y))to X&
&S(q)beta_Y:S(T(X)times T(Y))to Y
end{align}
is a product source in $mathscr D$.
Let
begin{align}
&u:Zto X&
&v:Zto Y
end{align}
be a source in $mathscr D$.
Then
begin{align}
&T(u):T(Z)to T(X)&
&T(v):T(Z)to T(Y)
end{align}
is a source in $mathscr C$, hence there exists one and only one morphism $w:T(Z)to T(X)times T(Y)$ such that
begin{align}
&T(u)=wp&
&T(v)=wq
end{align}
Then
begin{align}
beta^{-1}_ZS(w)S(p)beta_X&=beta^{-1}_Z(Scirc T)(u)beta_X&
beta^{-1}_ZS(w)S(q)beta_Y&=beta^{-1}_Z(Scirc T)(v)beta_Y\
&=beta^{-1}_Zbeta_Zu&
&=beta^{-1}_Zbeta_Zv\
&=u&
&=v
end{align}
which proves the existence part.
It remains to prove uniqueness.
For this, first recall that we can assume $alpha^{-1}_TT(beta)=1_T$ and $beta^{-1}_SS(alpha)=1_S$ (see, for example, here).
Let $w':Zto S(T(X)times T(Y))$ such that
begin{align}
&u=w'S(p)beta_X&
&v=w'S(q)beta_Y
end{align}
Then
begin{align}
T(u)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}p&
T(v)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}q\
&=T(w')alpha_{T(X)times T(Y)}p&
&=T(w')alpha_{T(X)times T(Y)}q\
end{align}
from which $w=T(w')alpha_{T(X)times T(Y)}$.
Consequently,
begin{align}
S(w)
&=beta_Zbeta^{-1}_Z(Scirc T)(w')S(alpha_{T(X)times T(Y)})\
&=beta_Zw'beta^{-1}_{S(T(X)times T(Y))}S(alpha_{T(X)times T(Y)})\
&=beta_Zw'
end{align}
thus concluding the proof.
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
Let $X,Y$ be objects in $mathscr D$ and
begin{align}
&p:T(X)times T(Y)to T(X)&
&q:T(X)times T(Y)to T(Y)
end{align}
be a product source in $mathscr C$.
We claim that
begin{align}
&S(p)beta_X:S(T(X)times T(Y))to X&
&S(q)beta_Y:S(T(X)times T(Y))to Y
end{align}
is a product source in $mathscr D$.
Let
begin{align}
&u:Zto X&
&v:Zto Y
end{align}
be a source in $mathscr D$.
Then
begin{align}
&T(u):T(Z)to T(X)&
&T(v):T(Z)to T(Y)
end{align}
is a source in $mathscr C$, hence there exists one and only one morphism $w:T(Z)to T(X)times T(Y)$ such that
begin{align}
&T(u)=wp&
&T(v)=wq
end{align}
Then
begin{align}
beta^{-1}_ZS(w)S(p)beta_X&=beta^{-1}_Z(Scirc T)(u)beta_X&
beta^{-1}_ZS(w)S(q)beta_Y&=beta^{-1}_Z(Scirc T)(v)beta_Y\
&=beta^{-1}_Zbeta_Zu&
&=beta^{-1}_Zbeta_Zv\
&=u&
&=v
end{align}
which proves the existence part.
It remains to prove uniqueness.
For this, first recall that we can assume $alpha^{-1}_TT(beta)=1_T$ and $beta^{-1}_SS(alpha)=1_S$ (see, for example, here).
Let $w':Zto S(T(X)times T(Y))$ such that
begin{align}
&u=w'S(p)beta_X&
&v=w'S(q)beta_Y
end{align}
Then
begin{align}
T(u)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}p&
T(v)&=alpha_{T(Z)}^{-1}T(beta_Z)T(w')alpha_{T(X)times T(Y)}q\
&=T(w')alpha_{T(X)times T(Y)}p&
&=T(w')alpha_{T(X)times T(Y)}q\
end{align}
from which $w=T(w')alpha_{T(X)times T(Y)}$.
Consequently,
begin{align}
S(w)
&=beta_Zbeta^{-1}_Z(Scirc T)(w')S(alpha_{T(X)times T(Y)})\
&=beta_Zw'beta^{-1}_{S(T(X)times T(Y))}S(alpha_{T(X)times T(Y)})\
&=beta_Zw'
end{align}
thus concluding the proof.
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
edited Nov 20 '18 at 22:49
darij grinberg
10.2k33062
10.2k33062
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
answered Nov 20 '18 at 11:03
Fabio Lucchini
7,83811426
7,83811426
Hmm. Your "source" means "cone in the particular case of a binary product", right? And you're writing composition the "other way round"?
– darij grinberg
Nov 20 '18 at 15:26
Yes, you are right;
– Fabio Lucchini
Nov 20 '18 at 16:13
@FabioLucchini I think beginning students should be warned about a choice like reversing the traditional composition order!
– Kevin Carlson
Nov 20 '18 at 17:26
add a comment |
Hmm. Your "source" means "cone in the particular case of a binary product", right? And you're writing composition the "other way round"?
– darij grinberg
Nov 20 '18 at 15:26
Yes, you are right;
– Fabio Lucchini
Nov 20 '18 at 16:13
@FabioLucchini I think beginning students should be warned about a choice like reversing the traditional composition order!
– Kevin Carlson
Nov 20 '18 at 17:26
Hmm. Your "source" means "cone in the particular case of a binary product", right? And you're writing composition the "other way round"?
– darij grinberg
Nov 20 '18 at 15:26
Hmm. Your "source" means "cone in the particular case of a binary product", right? And you're writing composition the "other way round"?
– darij grinberg
Nov 20 '18 at 15:26
Yes, you are right;
– Fabio Lucchini
Nov 20 '18 at 16:13
Yes, you are right;
– Fabio Lucchini
Nov 20 '18 at 16:13
@FabioLucchini I think beginning students should be warned about a choice like reversing the traditional composition order!
– Kevin Carlson
Nov 20 '18 at 17:26
@FabioLucchini I think beginning students should be warned about a choice like reversing the traditional composition order!
– Kevin Carlson
Nov 20 '18 at 17:26
add a comment |
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No, you can't just magically make arbitrary limits out of binary products.
– darij grinberg
Nov 20 '18 at 5:52
Okay then; what am I supposed to do in this?
– Eevee Trainer
Nov 20 '18 at 6:23
So you want to construct a product of two objects $X$ and $Y$ of $mathscr{D}$. You can construct binary products in $mathscr{C}$, so you can construct a product of $Tleft(Xright)$ and $Tleft(Yright)$. Furthermore, you can send this product back into $mathscr{D}$ using $S$. Does this do the trick?
– darij grinberg
Nov 20 '18 at 6:24
So $X,Y$ have image $T(X), T(Y)$ in $mathscr{C}$. Since $mathscr{C}$, by assumption, has binary products, then the object $T(X) times T(Y)$ must also exist in $mathscr{C}$. ... I'm a bit lost from there though. Obviously we can just send that product through $S$ back to $mathscr{D}$, but that doesn't really invoke any assumptions about the equivalence of the categories, namely the natural transformations (among other probable issues). So it's obviously more nuanced than that.
– Eevee Trainer
Nov 20 '18 at 6:48
@EeveeTrainer No, that's the right idea. The natural transformations go to show that $S$ sends the product to a product of $STX$ and $STY$, and thus of $X$ and $Y$.
– Kevin Carlson
Nov 20 '18 at 17:25