Mixing
Finding a colimit in the category of presheaves
$begingroup$
I have the following problem as part of my exam preporation and I need an idea how to approach it at least:
Let $mathfrak C$ be a small category and $F: mathfrak C^{opp} rightarrow mathfrak S et$ be a presheaf. Lets constract new category $mathfrak F$ as follows: $operatorname{Ob} mathfrak F = sqcup_{X in operatorname{Ob} mathfrak C} F(X) $ (disjoint union of all object images under the action of functor $F$) and for all $x in F(X)$ and $y in F(Y)$ $operatorname{Hom}_mathfrak F (x, y)$ consitst of all arrows $phi in operatorname{Hom}_{mathfrak C} (X, Y)$ such that $F phi (y) = x$.
Find colimit of the diagram $D: mathfrak F rightarrow mathfrak F unc(mathfrak C^{opp}, mathfrak Set)$ which maps object $x$ (which is also an element of $F(X)$) to presheaf $operatorname{Hom}_mathfrak C (-, X)$ and arrow $phi$ from $operatorname{Hom}_mathfrak F (x, y)$ to natural transformation $ phi_*$ of left multiplication on $phi$.
Thanks!
category-theory limits-colimits
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
$endgroup$
add a comment |
$begingroup$
I have the following problem as part of my exam preporation and I need an idea how to approach it at least:
Let $mathfrak C$ be a small category and $F: mathfrak C^{opp} rightarrow mathfrak S et$ be a presheaf. Lets constract new category $mathfrak F$ as follows: $operatorname{Ob} mathfrak F = sqcup_{X in operatorname{Ob} mathfrak C} F(X) $ (disjoint union of all object images under the action of functor $F$) and for all $x in F(X)$ and $y in F(Y)$ $operatorname{Hom}_mathfrak F (x, y)$ consitst of all arrows $phi in operatorname{Hom}_{mathfrak C} (X, Y)$ such that $F phi (y) = x$.
Find colimit of the diagram $D: mathfrak F rightarrow mathfrak F unc(mathfrak C^{opp}, mathfrak Set)$ which maps object $x$ (which is also an element of $F(X)$) to presheaf $operatorname{Hom}_mathfrak C (-, X)$ and arrow $phi$ from $operatorname{Hom}_mathfrak F (x, y)$ to natural transformation $ phi_*$ of left multiplication on $phi$.
Thanks!
category-theory limits-colimits
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
$endgroup$
$begingroup$
As a hint, think about the Yoneda lemma: the objects of $mathfrak{F}$ correspond to natural transformations $mathrm{Hom}(-,X)to F$, and the defining condition on $mathfrak{F}$'s hom-sets means for $f:(x,X)to (y,Y)$ in $mathfrak{F}$ , we have $alpha _{(y,Y)}hom(-,f)=alpha_{(x,X)}$ (where $alpha_{(x,X)}$ is the natural transformation $hom(-,X)to F$ corresponding to $xin F(x)$).
$endgroup$
– Malice Vidrine
Jan 20 at 17:26
$begingroup$
@MaliceVidrine May I ask you for more detailes? I still can't see what I have to do with it.
$endgroup$
– Gleb Chili
Jan 21 at 21:09
add a comment |
$begingroup$
I have the following problem as part of my exam preporation and I need an idea how to approach it at least:
Let $mathfrak C$ be a small category and $F: mathfrak C^{opp} rightarrow mathfrak S et$ be a presheaf. Lets constract new category $mathfrak F$ as follows: $operatorname{Ob} mathfrak F = sqcup_{X in operatorname{Ob} mathfrak C} F(X) $ (disjoint union of all object images under the action of functor $F$) and for all $x in F(X)$ and $y in F(Y)$ $operatorname{Hom}_mathfrak F (x, y)$ consitst of all arrows $phi in operatorname{Hom}_{mathfrak C} (X, Y)$ such that $F phi (y) = x$.
Find colimit of the diagram $D: mathfrak F rightarrow mathfrak F unc(mathfrak C^{opp}, mathfrak Set)$ which maps object $x$ (which is also an element of $F(X)$) to presheaf $operatorname{Hom}_mathfrak C (-, X)$ and arrow $phi$ from $operatorname{Hom}_mathfrak F (x, y)$ to natural transformation $ phi_*$ of left multiplication on $phi$.
Thanks!
category-theory limits-colimits
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
$endgroup$
I have the following problem as part of my exam preporation and I need an idea how to approach it at least:
Let $mathfrak C$ be a small category and $F: mathfrak C^{opp} rightarrow mathfrak S et$ be a presheaf. Lets constract new category $mathfrak F$ as follows: $operatorname{Ob} mathfrak F = sqcup_{X in operatorname{Ob} mathfrak C} F(X) $ (disjoint union of all object images under the action of functor $F$) and for all $x in F(X)$ and $y in F(Y)$ $operatorname{Hom}_mathfrak F (x, y)$ consitst of all arrows $phi in operatorname{Hom}_{mathfrak C} (X, Y)$ such that $F phi (y) = x$.
Find colimit of the diagram $D: mathfrak F rightarrow mathfrak F unc(mathfrak C^{opp}, mathfrak Set)$ which maps object $x$ (which is also an element of $F(X)$) to presheaf $operatorname{Hom}_mathfrak C (-, X)$ and arrow $phi$ from $operatorname{Hom}_mathfrak F (x, y)$ to natural transformation $ phi_*$ of left multiplication on $phi$.
Thanks!
category-theory limits-colimits
category-theory limits-colimits
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
asked Jan 20 at 17:01
Gleb ChiliGleb Chili
45428
45428
$begingroup$
As a hint, think about the Yoneda lemma: the objects of $mathfrak{F}$ correspond to natural transformations $mathrm{Hom}(-,X)to F$, and the defining condition on $mathfrak{F}$'s hom-sets means for $f:(x,X)to (y,Y)$ in $mathfrak{F}$ , we have $alpha _{(y,Y)}hom(-,f)=alpha_{(x,X)}$ (where $alpha_{(x,X)}$ is the natural transformation $hom(-,X)to F$ corresponding to $xin F(x)$).
$endgroup$
– Malice Vidrine
Jan 20 at 17:26
$begingroup$
@MaliceVidrine May I ask you for more detailes? I still can't see what I have to do with it.
$endgroup$
– Gleb Chili
Jan 21 at 21:09
add a comment |
$begingroup$
As a hint, think about the Yoneda lemma: the objects of $mathfrak{F}$ correspond to natural transformations $mathrm{Hom}(-,X)to F$, and the defining condition on $mathfrak{F}$'s hom-sets means for $f:(x,X)to (y,Y)$ in $mathfrak{F}$ , we have $alpha _{(y,Y)}hom(-,f)=alpha_{(x,X)}$ (where $alpha_{(x,X)}$ is the natural transformation $hom(-,X)to F$ corresponding to $xin F(x)$).
$endgroup$
– Malice Vidrine
Jan 20 at 17:26
$begingroup$
@MaliceVidrine May I ask you for more detailes? I still can't see what I have to do with it.
$endgroup$
– Gleb Chili
Jan 21 at 21:09
$begingroup$
As a hint, think about the Yoneda lemma: the objects of $mathfrak{F}$ correspond to natural transformations $mathrm{Hom}(-,X)to F$, and the defining condition on $mathfrak{F}$'s hom-sets means for $f:(x,X)to (y,Y)$ in $mathfrak{F}$ , we have $alpha _{(y,Y)}hom(-,f)=alpha_{(x,X)}$ (where $alpha_{(x,X)}$ is the natural transformation $hom(-,X)to F$ corresponding to $xin F(x)$).
$endgroup$
– Malice Vidrine
Jan 20 at 17:26
$begingroup$
As a hint, think about the Yoneda lemma: the objects of $mathfrak{F}$ correspond to natural transformations $mathrm{Hom}(-,X)to F$, and the defining condition on $mathfrak{F}$'s hom-sets means for $f:(x,X)to (y,Y)$ in $mathfrak{F}$ , we have $alpha _{(y,Y)}hom(-,f)=alpha_{(x,X)}$ (where $alpha_{(x,X)}$ is the natural transformation $hom(-,X)to F$ corresponding to $xin F(x)$).
$endgroup$
– Malice Vidrine
Jan 20 at 17:26
$begingroup$
@MaliceVidrine May I ask you for more detailes? I still can't see what I have to do with it.
$endgroup$
– Gleb Chili
Jan 21 at 21:09
$begingroup$
@MaliceVidrine May I ask you for more detailes? I still can't see what I have to do with it.
$endgroup$
– Gleb Chili
Jan 21 at 21:09
add a comment |
1 Answer
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oldest
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$begingroup$
It is going to turn out that the colimit of the diagram $D$ is just going to be $F$ itself, due to the Yoneda lemma.
We define a co-cone $alpha$ under $D$ by setting $alpha_{(x,X)}:D(x,X)to F$ to be the unique natural transformation $hom(-,X)to F$ that sends $id_X$ to $xin F(X)$. To show that this is natural, consider that for $f:(x,X)to (y,Y)$, we need $$alpha_{(y,Y)}circhom(-,f)=alpha_{(x,X)},$$ and by the Yoneda lemma it is sufficient to say that the left hand composite sends $id_X$ to $x$. First, $hom(X,f)(id_X)=finhom(X,Y)$; then because $$(alpha_{(y,Y)})_Y(id_Y)=y$$ it must be the case that $$(alpha_{(y,Y)})_X(f)=F(f)(y).$$ But since $f$ was a morphism $(x,X)to (y,Y)$ in $mathfrak{F}$, this means that $F(f)(y)=x$. This shows that $alpha$ is a co-cone as desired.
(Because it's easy to get lost in the subscripts, remember that $alpha_{(x,X)}$ is a natural transformation $hom(-,X)to F$, which has as its component at each $Yinmathfrak{C}^{op}$ the function $(alpha_{(x,X)})_Y:hom(Y,X)to F(Y)$.)
To show that this is a colimit, assume that you have a co-cone $beta$ under $D$ to another presheaf $G$. It will again be the case that each component $beta_{(x,X)}:hom(-,X)to G$ picks out a unique $$g_{x,X}=(beta_{(x,X)})_X(id_X)in G(X)$$ by the Yoneda lemma. What you wish to show is that the collection of functions ${gamma_X}_{Xinmathfrak{C}^{op}}$ defined by $$gamma_X(x)=g_{x,X}$$ defines a natural transformation $gamma:Fto G$, and does so uniquely. In this case, "uniquely" means that for each $(x,X)$, the morphisms $gamma_Xcircalpha_{(x,X)}$ are equal to $beta_{(x,X)}$; and conversely, given any natural transformation $chi:Fto G$, that the co-cone $Dto G$ whose components at $(x,X)$ are $chi_Xcircalpha_{(x,X)}$, simply gives back $chi$ when fed to the above construction. This is a mouthful, but it's actually quite simple (though not short) to show.
Hopefully this helps!
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
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add a comment |
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$begingroup$
It is going to turn out that the colimit of the diagram $D$ is just going to be $F$ itself, due to the Yoneda lemma.
We define a co-cone $alpha$ under $D$ by setting $alpha_{(x,X)}:D(x,X)to F$ to be the unique natural transformation $hom(-,X)to F$ that sends $id_X$ to $xin F(X)$. To show that this is natural, consider that for $f:(x,X)to (y,Y)$, we need $$alpha_{(y,Y)}circhom(-,f)=alpha_{(x,X)},$$ and by the Yoneda lemma it is sufficient to say that the left hand composite sends $id_X$ to $x$. First, $hom(X,f)(id_X)=finhom(X,Y)$; then because $$(alpha_{(y,Y)})_Y(id_Y)=y$$ it must be the case that $$(alpha_{(y,Y)})_X(f)=F(f)(y).$$ But since $f$ was a morphism $(x,X)to (y,Y)$ in $mathfrak{F}$, this means that $F(f)(y)=x$. This shows that $alpha$ is a co-cone as desired.
(Because it's easy to get lost in the subscripts, remember that $alpha_{(x,X)}$ is a natural transformation $hom(-,X)to F$, which has as its component at each $Yinmathfrak{C}^{op}$ the function $(alpha_{(x,X)})_Y:hom(Y,X)to F(Y)$.)
To show that this is a colimit, assume that you have a co-cone $beta$ under $D$ to another presheaf $G$. It will again be the case that each component $beta_{(x,X)}:hom(-,X)to G$ picks out a unique $$g_{x,X}=(beta_{(x,X)})_X(id_X)in G(X)$$ by the Yoneda lemma. What you wish to show is that the collection of functions ${gamma_X}_{Xinmathfrak{C}^{op}}$ defined by $$gamma_X(x)=g_{x,X}$$ defines a natural transformation $gamma:Fto G$, and does so uniquely. In this case, "uniquely" means that for each $(x,X)$, the morphisms $gamma_Xcircalpha_{(x,X)}$ are equal to $beta_{(x,X)}$; and conversely, given any natural transformation $chi:Fto G$, that the co-cone $Dto G$ whose components at $(x,X)$ are $chi_Xcircalpha_{(x,X)}$, simply gives back $chi$ when fed to the above construction. This is a mouthful, but it's actually quite simple (though not short) to show.
Hopefully this helps!
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
$endgroup$
add a comment |
$begingroup$
It is going to turn out that the colimit of the diagram $D$ is just going to be $F$ itself, due to the Yoneda lemma.
We define a co-cone $alpha$ under $D$ by setting $alpha_{(x,X)}:D(x,X)to F$ to be the unique natural transformation $hom(-,X)to F$ that sends $id_X$ to $xin F(X)$. To show that this is natural, consider that for $f:(x,X)to (y,Y)$, we need $$alpha_{(y,Y)}circhom(-,f)=alpha_{(x,X)},$$ and by the Yoneda lemma it is sufficient to say that the left hand composite sends $id_X$ to $x$. First, $hom(X,f)(id_X)=finhom(X,Y)$; then because $$(alpha_{(y,Y)})_Y(id_Y)=y$$ it must be the case that $$(alpha_{(y,Y)})_X(f)=F(f)(y).$$ But since $f$ was a morphism $(x,X)to (y,Y)$ in $mathfrak{F}$, this means that $F(f)(y)=x$. This shows that $alpha$ is a co-cone as desired.
(Because it's easy to get lost in the subscripts, remember that $alpha_{(x,X)}$ is a natural transformation $hom(-,X)to F$, which has as its component at each $Yinmathfrak{C}^{op}$ the function $(alpha_{(x,X)})_Y:hom(Y,X)to F(Y)$.)
To show that this is a colimit, assume that you have a co-cone $beta$ under $D$ to another presheaf $G$. It will again be the case that each component $beta_{(x,X)}:hom(-,X)to G$ picks out a unique $$g_{x,X}=(beta_{(x,X)})_X(id_X)in G(X)$$ by the Yoneda lemma. What you wish to show is that the collection of functions ${gamma_X}_{Xinmathfrak{C}^{op}}$ defined by $$gamma_X(x)=g_{x,X}$$ defines a natural transformation $gamma:Fto G$, and does so uniquely. In this case, "uniquely" means that for each $(x,X)$, the morphisms $gamma_Xcircalpha_{(x,X)}$ are equal to $beta_{(x,X)}$; and conversely, given any natural transformation $chi:Fto G$, that the co-cone $Dto G$ whose components at $(x,X)$ are $chi_Xcircalpha_{(x,X)}$, simply gives back $chi$ when fed to the above construction. This is a mouthful, but it's actually quite simple (though not short) to show.
Hopefully this helps!
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
$endgroup$
add a comment |
$begingroup$
It is going to turn out that the colimit of the diagram $D$ is just going to be $F$ itself, due to the Yoneda lemma.
We define a co-cone $alpha$ under $D$ by setting $alpha_{(x,X)}:D(x,X)to F$ to be the unique natural transformation $hom(-,X)to F$ that sends $id_X$ to $xin F(X)$. To show that this is natural, consider that for $f:(x,X)to (y,Y)$, we need $$alpha_{(y,Y)}circhom(-,f)=alpha_{(x,X)},$$ and by the Yoneda lemma it is sufficient to say that the left hand composite sends $id_X$ to $x$. First, $hom(X,f)(id_X)=finhom(X,Y)$; then because $$(alpha_{(y,Y)})_Y(id_Y)=y$$ it must be the case that $$(alpha_{(y,Y)})_X(f)=F(f)(y).$$ But since $f$ was a morphism $(x,X)to (y,Y)$ in $mathfrak{F}$, this means that $F(f)(y)=x$. This shows that $alpha$ is a co-cone as desired.
(Because it's easy to get lost in the subscripts, remember that $alpha_{(x,X)}$ is a natural transformation $hom(-,X)to F$, which has as its component at each $Yinmathfrak{C}^{op}$ the function $(alpha_{(x,X)})_Y:hom(Y,X)to F(Y)$.)
To show that this is a colimit, assume that you have a co-cone $beta$ under $D$ to another presheaf $G$. It will again be the case that each component $beta_{(x,X)}:hom(-,X)to G$ picks out a unique $$g_{x,X}=(beta_{(x,X)})_X(id_X)in G(X)$$ by the Yoneda lemma. What you wish to show is that the collection of functions ${gamma_X}_{Xinmathfrak{C}^{op}}$ defined by $$gamma_X(x)=g_{x,X}$$ defines a natural transformation $gamma:Fto G$, and does so uniquely. In this case, "uniquely" means that for each $(x,X)$, the morphisms $gamma_Xcircalpha_{(x,X)}$ are equal to $beta_{(x,X)}$; and conversely, given any natural transformation $chi:Fto G$, that the co-cone $Dto G$ whose components at $(x,X)$ are $chi_Xcircalpha_{(x,X)}$, simply gives back $chi$ when fed to the above construction. This is a mouthful, but it's actually quite simple (though not short) to show.
Hopefully this helps!
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
$endgroup$
It is going to turn out that the colimit of the diagram $D$ is just going to be $F$ itself, due to the Yoneda lemma.
We define a co-cone $alpha$ under $D$ by setting $alpha_{(x,X)}:D(x,X)to F$ to be the unique natural transformation $hom(-,X)to F$ that sends $id_X$ to $xin F(X)$. To show that this is natural, consider that for $f:(x,X)to (y,Y)$, we need $$alpha_{(y,Y)}circhom(-,f)=alpha_{(x,X)},$$ and by the Yoneda lemma it is sufficient to say that the left hand composite sends $id_X$ to $x$. First, $hom(X,f)(id_X)=finhom(X,Y)$; then because $$(alpha_{(y,Y)})_Y(id_Y)=y$$ it must be the case that $$(alpha_{(y,Y)})_X(f)=F(f)(y).$$ But since $f$ was a morphism $(x,X)to (y,Y)$ in $mathfrak{F}$, this means that $F(f)(y)=x$. This shows that $alpha$ is a co-cone as desired.
(Because it's easy to get lost in the subscripts, remember that $alpha_{(x,X)}$ is a natural transformation $hom(-,X)to F$, which has as its component at each $Yinmathfrak{C}^{op}$ the function $(alpha_{(x,X)})_Y:hom(Y,X)to F(Y)$.)
To show that this is a colimit, assume that you have a co-cone $beta$ under $D$ to another presheaf $G$. It will again be the case that each component $beta_{(x,X)}:hom(-,X)to G$ picks out a unique $$g_{x,X}=(beta_{(x,X)})_X(id_X)in G(X)$$ by the Yoneda lemma. What you wish to show is that the collection of functions ${gamma_X}_{Xinmathfrak{C}^{op}}$ defined by $$gamma_X(x)=g_{x,X}$$ defines a natural transformation $gamma:Fto G$, and does so uniquely. In this case, "uniquely" means that for each $(x,X)$, the morphisms $gamma_Xcircalpha_{(x,X)}$ are equal to $beta_{(x,X)}$; and conversely, given any natural transformation $chi:Fto G$, that the co-cone $Dto G$ whose components at $(x,X)$ are $chi_Xcircalpha_{(x,X)}$, simply gives back $chi$ when fed to the above construction. This is a mouthful, but it's actually quite simple (though not short) to show.
Hopefully this helps!
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
edited Jan 21 at 22:45
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
answered Jan 21 at 22:18
Malice VidrineMalice Vidrine
6,17921123
6,17921123
add a comment |
add a comment |
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$begingroup$
As a hint, think about the Yoneda lemma: the objects of $mathfrak{F}$ correspond to natural transformations $mathrm{Hom}(-,X)to F$, and the defining condition on $mathfrak{F}$'s hom-sets means for $f:(x,X)to (y,Y)$ in $mathfrak{F}$ , we have $alpha _{(y,Y)}hom(-,f)=alpha_{(x,X)}$ (where $alpha_{(x,X)}$ is the natural transformation $hom(-,X)to F$ corresponding to $xin F(x)$).
$endgroup$
– Malice Vidrine
Jan 20 at 17:26
$begingroup$
@MaliceVidrine May I ask you for more detailes? I still can't see what I have to do with it.
$endgroup$
– Gleb Chili
Jan 21 at 21:09