Hom, Limits and Colimits












0












$begingroup$


Let $I$ be a small category and $C$ a category. Suppose that
$$ Hom_C ( A, lim_{ leftarrow I} D_i) cong lim_{leftarrow I} Hom(A,D_i) $$ for $A in C$ and $D: I rightarrow C$ when the expression above makes sense. Is it true that for the same $A$ and $D$ we have
$$ Hom ( lim_{ rightarrow I} D_i, A) cong lim_{leftarrow I} Hom(D_i,A)?$$



I was wondering if this is true, but I couldn't figured out a solution. Could someone give me a hint or an answer?






category-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    These are true for all $A$ and $D$ (if the limit or colimit of $D$ exists) by definition of (co)limit.
    $endgroup$
    – Arnaud D.
    Nov 29 '18 at 13:57


















0












$begingroup$


Let $I$ be a small category and $C$ a category. Suppose that
$$ Hom_C ( A, lim_{ leftarrow I} D_i) cong lim_{leftarrow I} Hom(A,D_i) $$ for $A in C$ and $D: I rightarrow C$ when the expression above makes sense. Is it true that for the same $A$ and $D$ we have
$$ Hom ( lim_{ rightarrow I} D_i, A) cong lim_{leftarrow I} Hom(D_i,A)?$$



I was wondering if this is true, but I couldn't figured out a solution. Could someone give me a hint or an answer?






category-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    These are true for all $A$ and $D$ (if the limit or colimit of $D$ exists) by definition of (co)limit.
    $endgroup$
    – Arnaud D.
    Nov 29 '18 at 13:57
















0












0








0





$begingroup$


Let $I$ be a small category and $C$ a category. Suppose that
$$ Hom_C ( A, lim_{ leftarrow I} D_i) cong lim_{leftarrow I} Hom(A,D_i) $$ for $A in C$ and $D: I rightarrow C$ when the expression above makes sense. Is it true that for the same $A$ and $D$ we have
$$ Hom ( lim_{ rightarrow I} D_i, A) cong lim_{leftarrow I} Hom(D_i,A)?$$



I was wondering if this is true, but I couldn't figured out a solution. Could someone give me a hint or an answer?






category-theory







share|cite|improve this question









$endgroup$




Let $I$ be a small category and $C$ a category. Suppose that
$$ Hom_C ( A, lim_{ leftarrow I} D_i) cong lim_{leftarrow I} Hom(A,D_i) $$ for $A in C$ and $D: I rightarrow C$ when the expression above makes sense. Is it true that for the same $A$ and $D$ we have
$$ Hom ( lim_{ rightarrow I} D_i, A) cong lim_{leftarrow I} Hom(D_i,A)?$$



I was wondering if this is true, but I couldn't figured out a solution. Could someone give me a hint or an answer?






category-theory




category-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 13:46









H RH R

1608




1608












  • $begingroup$
    These are true for all $A$ and $D$ (if the limit or colimit of $D$ exists) by definition of (co)limit.
    $endgroup$
    – Arnaud D.
    Nov 29 '18 at 13:57




















  • $begingroup$
    These are true for all $A$ and $D$ (if the limit or colimit of $D$ exists) by definition of (co)limit.
    $endgroup$
    – Arnaud D.
    Nov 29 '18 at 13:57


















$begingroup$
These are true for all $A$ and $D$ (if the limit or colimit of $D$ exists) by definition of (co)limit.
$endgroup$
– Arnaud D.
Nov 29 '18 at 13:57






$begingroup$
These are true for all $A$ and $D$ (if the limit or colimit of $D$ exists) by definition of (co)limit.
$endgroup$
– Arnaud D.
Nov 29 '18 at 13:57












1 Answer
1






active

oldest

votes


















2












$begingroup$

First off: both your statements, given the existence of the (co)limits in question, are true and easy to find in standard literature. Furthermore, assuming that we only know the first statement, we can conclude the second by applying the first to the opposite functor $D^text{op}: I^text{op} to D^text{op}$:



Taking the opposite category converts limits into colimits and vice versa. I.e., denoting the same objects in the opposite category with $D_i^text{op}$, we have $lim_{rightarrow I^text{op}}(D_i^text{op}) = (lim_{leftarrow I} D_i)^text{op}$, or in more modern notation, $lim D^text{op} = (text{colim} D)^text{op}$, where by $(lim D)^text{op}$ I mean the corresponding cocone in the opposite category. Thus the second isomorphism is derived as follows:



$$
Hom_C ( lim_{ rightarrow I} D_i, A) = Hom_{C^text{op}}(A, lim_{ leftarrow I^text{op}} D_i^text{op}) cong lim_{ leftarrow I^text{op}} Hom_{C^text{op}}(A, D_i^text{op}) = lim_{leftarrow I} Hom_C(D_i,A)
$$



So no, existence of $lim D$ does not imply existence of $text{colim} D$, but only of $text{colim} D^text{op}$. But if it exists, this shows that your second isomorphism directly follows from the first.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the existance of $lim_{leftarrow I} D_i$ implies the existance of $lim_{rightarrow I} D_i$?
    $endgroup$
    – H R
    Nov 29 '18 at 16:07












  • $begingroup$
    No, it doen't. I'll extend my answer soon.
    $endgroup$
    – Gnampfissimo
    Nov 29 '18 at 17:37











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1 Answer
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1 Answer
1






active

oldest

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active

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votes









2












$begingroup$

First off: both your statements, given the existence of the (co)limits in question, are true and easy to find in standard literature. Furthermore, assuming that we only know the first statement, we can conclude the second by applying the first to the opposite functor $D^text{op}: I^text{op} to D^text{op}$:



Taking the opposite category converts limits into colimits and vice versa. I.e., denoting the same objects in the opposite category with $D_i^text{op}$, we have $lim_{rightarrow I^text{op}}(D_i^text{op}) = (lim_{leftarrow I} D_i)^text{op}$, or in more modern notation, $lim D^text{op} = (text{colim} D)^text{op}$, where by $(lim D)^text{op}$ I mean the corresponding cocone in the opposite category. Thus the second isomorphism is derived as follows:



$$
Hom_C ( lim_{ rightarrow I} D_i, A) = Hom_{C^text{op}}(A, lim_{ leftarrow I^text{op}} D_i^text{op}) cong lim_{ leftarrow I^text{op}} Hom_{C^text{op}}(A, D_i^text{op}) = lim_{leftarrow I} Hom_C(D_i,A)
$$



So no, existence of $lim D$ does not imply existence of $text{colim} D$, but only of $text{colim} D^text{op}$. But if it exists, this shows that your second isomorphism directly follows from the first.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the existance of $lim_{leftarrow I} D_i$ implies the existance of $lim_{rightarrow I} D_i$?
    $endgroup$
    – H R
    Nov 29 '18 at 16:07












  • $begingroup$
    No, it doen't. I'll extend my answer soon.
    $endgroup$
    – Gnampfissimo
    Nov 29 '18 at 17:37
















2












$begingroup$

First off: both your statements, given the existence of the (co)limits in question, are true and easy to find in standard literature. Furthermore, assuming that we only know the first statement, we can conclude the second by applying the first to the opposite functor $D^text{op}: I^text{op} to D^text{op}$:



Taking the opposite category converts limits into colimits and vice versa. I.e., denoting the same objects in the opposite category with $D_i^text{op}$, we have $lim_{rightarrow I^text{op}}(D_i^text{op}) = (lim_{leftarrow I} D_i)^text{op}$, or in more modern notation, $lim D^text{op} = (text{colim} D)^text{op}$, where by $(lim D)^text{op}$ I mean the corresponding cocone in the opposite category. Thus the second isomorphism is derived as follows:



$$
Hom_C ( lim_{ rightarrow I} D_i, A) = Hom_{C^text{op}}(A, lim_{ leftarrow I^text{op}} D_i^text{op}) cong lim_{ leftarrow I^text{op}} Hom_{C^text{op}}(A, D_i^text{op}) = lim_{leftarrow I} Hom_C(D_i,A)
$$



So no, existence of $lim D$ does not imply existence of $text{colim} D$, but only of $text{colim} D^text{op}$. But if it exists, this shows that your second isomorphism directly follows from the first.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the existance of $lim_{leftarrow I} D_i$ implies the existance of $lim_{rightarrow I} D_i$?
    $endgroup$
    – H R
    Nov 29 '18 at 16:07












  • $begingroup$
    No, it doen't. I'll extend my answer soon.
    $endgroup$
    – Gnampfissimo
    Nov 29 '18 at 17:37














2












2








2





$begingroup$

First off: both your statements, given the existence of the (co)limits in question, are true and easy to find in standard literature. Furthermore, assuming that we only know the first statement, we can conclude the second by applying the first to the opposite functor $D^text{op}: I^text{op} to D^text{op}$:



Taking the opposite category converts limits into colimits and vice versa. I.e., denoting the same objects in the opposite category with $D_i^text{op}$, we have $lim_{rightarrow I^text{op}}(D_i^text{op}) = (lim_{leftarrow I} D_i)^text{op}$, or in more modern notation, $lim D^text{op} = (text{colim} D)^text{op}$, where by $(lim D)^text{op}$ I mean the corresponding cocone in the opposite category. Thus the second isomorphism is derived as follows:



$$
Hom_C ( lim_{ rightarrow I} D_i, A) = Hom_{C^text{op}}(A, lim_{ leftarrow I^text{op}} D_i^text{op}) cong lim_{ leftarrow I^text{op}} Hom_{C^text{op}}(A, D_i^text{op}) = lim_{leftarrow I} Hom_C(D_i,A)
$$



So no, existence of $lim D$ does not imply existence of $text{colim} D$, but only of $text{colim} D^text{op}$. But if it exists, this shows that your second isomorphism directly follows from the first.






share|cite|improve this answer











$endgroup$



First off: both your statements, given the existence of the (co)limits in question, are true and easy to find in standard literature. Furthermore, assuming that we only know the first statement, we can conclude the second by applying the first to the opposite functor $D^text{op}: I^text{op} to D^text{op}$:



Taking the opposite category converts limits into colimits and vice versa. I.e., denoting the same objects in the opposite category with $D_i^text{op}$, we have $lim_{rightarrow I^text{op}}(D_i^text{op}) = (lim_{leftarrow I} D_i)^text{op}$, or in more modern notation, $lim D^text{op} = (text{colim} D)^text{op}$, where by $(lim D)^text{op}$ I mean the corresponding cocone in the opposite category. Thus the second isomorphism is derived as follows:



$$
Hom_C ( lim_{ rightarrow I} D_i, A) = Hom_{C^text{op}}(A, lim_{ leftarrow I^text{op}} D_i^text{op}) cong lim_{ leftarrow I^text{op}} Hom_{C^text{op}}(A, D_i^text{op}) = lim_{leftarrow I} Hom_C(D_i,A)
$$



So no, existence of $lim D$ does not imply existence of $text{colim} D$, but only of $text{colim} D^text{op}$. But if it exists, this shows that your second isomorphism directly follows from the first.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 18:20

























answered Nov 29 '18 at 13:57









GnampfissimoGnampfissimo

18011




18011












  • $begingroup$
    But the existance of $lim_{leftarrow I} D_i$ implies the existance of $lim_{rightarrow I} D_i$?
    $endgroup$
    – H R
    Nov 29 '18 at 16:07












  • $begingroup$
    No, it doen't. I'll extend my answer soon.
    $endgroup$
    – Gnampfissimo
    Nov 29 '18 at 17:37


















  • $begingroup$
    But the existance of $lim_{leftarrow I} D_i$ implies the existance of $lim_{rightarrow I} D_i$?
    $endgroup$
    – H R
    Nov 29 '18 at 16:07












  • $begingroup$
    No, it doen't. I'll extend my answer soon.
    $endgroup$
    – Gnampfissimo
    Nov 29 '18 at 17:37
















$begingroup$
But the existance of $lim_{leftarrow I} D_i$ implies the existance of $lim_{rightarrow I} D_i$?
$endgroup$
– H R
Nov 29 '18 at 16:07






$begingroup$
But the existance of $lim_{leftarrow I} D_i$ implies the existance of $lim_{rightarrow I} D_i$?
$endgroup$
– H R
Nov 29 '18 at 16:07














$begingroup$
No, it doen't. I'll extend my answer soon.
$endgroup$
– Gnampfissimo
Nov 29 '18 at 17:37




$begingroup$
No, it doen't. I'll extend my answer soon.
$endgroup$
– Gnampfissimo
Nov 29 '18 at 17:37


















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