Category with colimits but no limits












4












$begingroup$


(I suspect this is a very easy question: I haven't spent much time thinking about category theory.) $DeclareMathOperator{colim}{colim}DeclareMathOperator{Dom}{Dom}DeclareMathOperator{im}{im}$



In The Rising Sea, Ravi Vakil notes that our intuition for limits and colimits of the diagram $$lim{X_j}todotsto X_{-2}to X_{-1}to X_0to X_1to X_2todotstocolim{X_j}$$ (the dots may be finite and diagonal morphisms were omitted to fit within MathJax) are very different. To wit, each element of $lim{X_j}$ is a sequence of "compatible" elements from the ${X_j}_j$, whereas each element of $colim{X_j}$ is a single distinguished element from $X_k$.



In my mind, this discrepancy occurs because of a fundamental asymmetry in the intuition regarding homomorphism functors. To wit, if we have $fin X_0to X_1$, then we automatically assume that there will be some remnant of $X_0$ in $im{(f)}$. Conversely, given $fin X_{-1}to X_0$, we do not make the assumption that $Dom{(f)}=X_{-1}$ contains all the properties of $X_0$; some may instead be "emergent."



To put it another way, it seems harder to construct domains than ranges for functions. Is this intuition rooted in truth; that is, does there exist a category with colimits but no limits?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    There is no category with "no limits"; however there are categories that have "more" colimits than limits
    $endgroup$
    – Max
    6 hours ago










  • $begingroup$
    @Max: That statement, along with an example of such a category, sounds like a complete answer I'd accept...*hint, hint*
    $endgroup$
    – Jacob Manaker
    6 hours ago






  • 1




    $begingroup$
    I don't agree with your "that is" at all: your specific question at the end has almost nothing to do with whether your intuition is rooted in truth.
    $endgroup$
    – Eric Wofsey
    6 hours ago






  • 2




    $begingroup$
    @Max: How about the empty category? :) Of course, that doesn't have any colimits either.
    $endgroup$
    – Eric Wofsey
    6 hours ago












  • $begingroup$
    @EricWofsey : woops always forget that one; adding "nontrivial" should solve it (it always does)
    $endgroup$
    – Max
    6 hours ago
















4












$begingroup$


(I suspect this is a very easy question: I haven't spent much time thinking about category theory.) $DeclareMathOperator{colim}{colim}DeclareMathOperator{Dom}{Dom}DeclareMathOperator{im}{im}$



In The Rising Sea, Ravi Vakil notes that our intuition for limits and colimits of the diagram $$lim{X_j}todotsto X_{-2}to X_{-1}to X_0to X_1to X_2todotstocolim{X_j}$$ (the dots may be finite and diagonal morphisms were omitted to fit within MathJax) are very different. To wit, each element of $lim{X_j}$ is a sequence of "compatible" elements from the ${X_j}_j$, whereas each element of $colim{X_j}$ is a single distinguished element from $X_k$.



In my mind, this discrepancy occurs because of a fundamental asymmetry in the intuition regarding homomorphism functors. To wit, if we have $fin X_0to X_1$, then we automatically assume that there will be some remnant of $X_0$ in $im{(f)}$. Conversely, given $fin X_{-1}to X_0$, we do not make the assumption that $Dom{(f)}=X_{-1}$ contains all the properties of $X_0$; some may instead be "emergent."



To put it another way, it seems harder to construct domains than ranges for functions. Is this intuition rooted in truth; that is, does there exist a category with colimits but no limits?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    There is no category with "no limits"; however there are categories that have "more" colimits than limits
    $endgroup$
    – Max
    6 hours ago










  • $begingroup$
    @Max: That statement, along with an example of such a category, sounds like a complete answer I'd accept...*hint, hint*
    $endgroup$
    – Jacob Manaker
    6 hours ago






  • 1




    $begingroup$
    I don't agree with your "that is" at all: your specific question at the end has almost nothing to do with whether your intuition is rooted in truth.
    $endgroup$
    – Eric Wofsey
    6 hours ago






  • 2




    $begingroup$
    @Max: How about the empty category? :) Of course, that doesn't have any colimits either.
    $endgroup$
    – Eric Wofsey
    6 hours ago












  • $begingroup$
    @EricWofsey : woops always forget that one; adding "nontrivial" should solve it (it always does)
    $endgroup$
    – Max
    6 hours ago














4












4








4





$begingroup$


(I suspect this is a very easy question: I haven't spent much time thinking about category theory.) $DeclareMathOperator{colim}{colim}DeclareMathOperator{Dom}{Dom}DeclareMathOperator{im}{im}$



In The Rising Sea, Ravi Vakil notes that our intuition for limits and colimits of the diagram $$lim{X_j}todotsto X_{-2}to X_{-1}to X_0to X_1to X_2todotstocolim{X_j}$$ (the dots may be finite and diagonal morphisms were omitted to fit within MathJax) are very different. To wit, each element of $lim{X_j}$ is a sequence of "compatible" elements from the ${X_j}_j$, whereas each element of $colim{X_j}$ is a single distinguished element from $X_k$.



In my mind, this discrepancy occurs because of a fundamental asymmetry in the intuition regarding homomorphism functors. To wit, if we have $fin X_0to X_1$, then we automatically assume that there will be some remnant of $X_0$ in $im{(f)}$. Conversely, given $fin X_{-1}to X_0$, we do not make the assumption that $Dom{(f)}=X_{-1}$ contains all the properties of $X_0$; some may instead be "emergent."



To put it another way, it seems harder to construct domains than ranges for functions. Is this intuition rooted in truth; that is, does there exist a category with colimits but no limits?










share|cite|improve this question









$endgroup$




(I suspect this is a very easy question: I haven't spent much time thinking about category theory.) $DeclareMathOperator{colim}{colim}DeclareMathOperator{Dom}{Dom}DeclareMathOperator{im}{im}$



In The Rising Sea, Ravi Vakil notes that our intuition for limits and colimits of the diagram $$lim{X_j}todotsto X_{-2}to X_{-1}to X_0to X_1to X_2todotstocolim{X_j}$$ (the dots may be finite and diagonal morphisms were omitted to fit within MathJax) are very different. To wit, each element of $lim{X_j}$ is a sequence of "compatible" elements from the ${X_j}_j$, whereas each element of $colim{X_j}$ is a single distinguished element from $X_k$.



In my mind, this discrepancy occurs because of a fundamental asymmetry in the intuition regarding homomorphism functors. To wit, if we have $fin X_0to X_1$, then we automatically assume that there will be some remnant of $X_0$ in $im{(f)}$. Conversely, given $fin X_{-1}to X_0$, we do not make the assumption that $Dom{(f)}=X_{-1}$ contains all the properties of $X_0$; some may instead be "emergent."



To put it another way, it seems harder to construct domains than ranges for functions. Is this intuition rooted in truth; that is, does there exist a category with colimits but no limits?







category-theory intuition limits-colimits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 6 hours ago









Jacob ManakerJacob Manaker

1,221416




1,221416








  • 1




    $begingroup$
    There is no category with "no limits"; however there are categories that have "more" colimits than limits
    $endgroup$
    – Max
    6 hours ago










  • $begingroup$
    @Max: That statement, along with an example of such a category, sounds like a complete answer I'd accept...*hint, hint*
    $endgroup$
    – Jacob Manaker
    6 hours ago






  • 1




    $begingroup$
    I don't agree with your "that is" at all: your specific question at the end has almost nothing to do with whether your intuition is rooted in truth.
    $endgroup$
    – Eric Wofsey
    6 hours ago






  • 2




    $begingroup$
    @Max: How about the empty category? :) Of course, that doesn't have any colimits either.
    $endgroup$
    – Eric Wofsey
    6 hours ago












  • $begingroup$
    @EricWofsey : woops always forget that one; adding "nontrivial" should solve it (it always does)
    $endgroup$
    – Max
    6 hours ago














  • 1




    $begingroup$
    There is no category with "no limits"; however there are categories that have "more" colimits than limits
    $endgroup$
    – Max
    6 hours ago










  • $begingroup$
    @Max: That statement, along with an example of such a category, sounds like a complete answer I'd accept...*hint, hint*
    $endgroup$
    – Jacob Manaker
    6 hours ago






  • 1




    $begingroup$
    I don't agree with your "that is" at all: your specific question at the end has almost nothing to do with whether your intuition is rooted in truth.
    $endgroup$
    – Eric Wofsey
    6 hours ago






  • 2




    $begingroup$
    @Max: How about the empty category? :) Of course, that doesn't have any colimits either.
    $endgroup$
    – Eric Wofsey
    6 hours ago












  • $begingroup$
    @EricWofsey : woops always forget that one; adding "nontrivial" should solve it (it always does)
    $endgroup$
    – Max
    6 hours ago








1




1




$begingroup$
There is no category with "no limits"; however there are categories that have "more" colimits than limits
$endgroup$
– Max
6 hours ago




$begingroup$
There is no category with "no limits"; however there are categories that have "more" colimits than limits
$endgroup$
– Max
6 hours ago












$begingroup$
@Max: That statement, along with an example of such a category, sounds like a complete answer I'd accept...*hint, hint*
$endgroup$
– Jacob Manaker
6 hours ago




$begingroup$
@Max: That statement, along with an example of such a category, sounds like a complete answer I'd accept...*hint, hint*
$endgroup$
– Jacob Manaker
6 hours ago




1




1




$begingroup$
I don't agree with your "that is" at all: your specific question at the end has almost nothing to do with whether your intuition is rooted in truth.
$endgroup$
– Eric Wofsey
6 hours ago




$begingroup$
I don't agree with your "that is" at all: your specific question at the end has almost nothing to do with whether your intuition is rooted in truth.
$endgroup$
– Eric Wofsey
6 hours ago




2




2




$begingroup$
@Max: How about the empty category? :) Of course, that doesn't have any colimits either.
$endgroup$
– Eric Wofsey
6 hours ago






$begingroup$
@Max: How about the empty category? :) Of course, that doesn't have any colimits either.
$endgroup$
– Eric Wofsey
6 hours ago














$begingroup$
@EricWofsey : woops always forget that one; adding "nontrivial" should solve it (it always does)
$endgroup$
– Max
6 hours ago




$begingroup$
@EricWofsey : woops always forget that one; adding "nontrivial" should solve it (it always does)
$endgroup$
– Max
6 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Sure, take for example the category $$A rightarrow B leftarrow C$$



Other than the identities, there is exactly one arrow from $A$ to $B$ and one arrow from $C$ to $B$.



It has all colimits -- the colimit of every diagram that contains two different objects is $B$, and of a diagram that contains only one object it is that object itself.



But it doesn't have all limits; for example there is no product of $A$ and $C$.





Of course, the opposite category of a category with all colimits and not all limits will be a category with all limits and not all colimits ...






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets.
    $endgroup$
    – Christoph
    6 hours ago



















3












$begingroup$

I object to the premise of your question, namely that the existence of a category with colimits but not limits has anything to do with the intuition you discuss. Note that in an arbitrary category, morphisms need not have any relation to functions. Moreover, given any category $C$, you can form the opposite category $C^{op}$ which reverses the directions of all the arrows. So, domains and codomains are really completely symmetric if you're just talking about arbitrary categories. For instance, if $C$ has all colimits but not all limits, then $C^{op}$ has all limits but not all colimits.



The intuition that Ravi Vakil is talking about is not applicable to arbitrary categories. Instead, it's applicable to the sort of "concrete" categories we tend to encounter in practice (but not their opposite categories), where morphisms are closely related to functions and limits and filtered colimits are constructed similarly to how they are in the category of sets.



As for the way you interpreted the intuition, I'm not entirely sure what you mean but it seems not entirely unreasonable. The way I might phrase it is that a map $X_nto X$ gives you actual elements of $X$ for each element of $X_0$. So, if $X$ is the colimit of the sequence, it has specific elements which come from the elements of each $X_n$ (and it can be deduced from the universal property that every element of $X$ comes from some $X_n$, at least in the category of sets). On the other hand, a map $Xto X_n$ doesn't tell you any specific elements of $X$. So, you can't describe what an element of $X$ can look like using just a single $X_n$; you really need the whole sequence before you can even name a single element of $X$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Sure, take for example the category $$A rightarrow B leftarrow C$$



    Other than the identities, there is exactly one arrow from $A$ to $B$ and one arrow from $C$ to $B$.



    It has all colimits -- the colimit of every diagram that contains two different objects is $B$, and of a diagram that contains only one object it is that object itself.



    But it doesn't have all limits; for example there is no product of $A$ and $C$.





    Of course, the opposite category of a category with all colimits and not all limits will be a category with all limits and not all colimits ...






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets.
      $endgroup$
      – Christoph
      6 hours ago
















    3












    $begingroup$

    Sure, take for example the category $$A rightarrow B leftarrow C$$



    Other than the identities, there is exactly one arrow from $A$ to $B$ and one arrow from $C$ to $B$.



    It has all colimits -- the colimit of every diagram that contains two different objects is $B$, and of a diagram that contains only one object it is that object itself.



    But it doesn't have all limits; for example there is no product of $A$ and $C$.





    Of course, the opposite category of a category with all colimits and not all limits will be a category with all limits and not all colimits ...






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets.
      $endgroup$
      – Christoph
      6 hours ago














    3












    3








    3





    $begingroup$

    Sure, take for example the category $$A rightarrow B leftarrow C$$



    Other than the identities, there is exactly one arrow from $A$ to $B$ and one arrow from $C$ to $B$.



    It has all colimits -- the colimit of every diagram that contains two different objects is $B$, and of a diagram that contains only one object it is that object itself.



    But it doesn't have all limits; for example there is no product of $A$ and $C$.





    Of course, the opposite category of a category with all colimits and not all limits will be a category with all limits and not all colimits ...






    share|cite|improve this answer











    $endgroup$



    Sure, take for example the category $$A rightarrow B leftarrow C$$



    Other than the identities, there is exactly one arrow from $A$ to $B$ and one arrow from $C$ to $B$.



    It has all colimits -- the colimit of every diagram that contains two different objects is $B$, and of a diagram that contains only one object it is that object itself.



    But it doesn't have all limits; for example there is no product of $A$ and $C$.





    Of course, the opposite category of a category with all colimits and not all limits will be a category with all limits and not all colimits ...







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 6 hours ago









    Joseph Martin

    412214




    412214










    answered 6 hours ago









    Henning MakholmHenning Makholm

    240k17306544




    240k17306544








    • 2




      $begingroup$
      Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets.
      $endgroup$
      – Christoph
      6 hours ago














    • 2




      $begingroup$
      Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets.
      $endgroup$
      – Christoph
      6 hours ago








    2




    2




    $begingroup$
    Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets.
    $endgroup$
    – Christoph
    6 hours ago




    $begingroup$
    Note that this example is a poset, where limits and colimits are just meets and joins. I add this since posets of form great examples and it's nice to think about what some fancy category theory terms mean for posets.
    $endgroup$
    – Christoph
    6 hours ago











    3












    $begingroup$

    I object to the premise of your question, namely that the existence of a category with colimits but not limits has anything to do with the intuition you discuss. Note that in an arbitrary category, morphisms need not have any relation to functions. Moreover, given any category $C$, you can form the opposite category $C^{op}$ which reverses the directions of all the arrows. So, domains and codomains are really completely symmetric if you're just talking about arbitrary categories. For instance, if $C$ has all colimits but not all limits, then $C^{op}$ has all limits but not all colimits.



    The intuition that Ravi Vakil is talking about is not applicable to arbitrary categories. Instead, it's applicable to the sort of "concrete" categories we tend to encounter in practice (but not their opposite categories), where morphisms are closely related to functions and limits and filtered colimits are constructed similarly to how they are in the category of sets.



    As for the way you interpreted the intuition, I'm not entirely sure what you mean but it seems not entirely unreasonable. The way I might phrase it is that a map $X_nto X$ gives you actual elements of $X$ for each element of $X_0$. So, if $X$ is the colimit of the sequence, it has specific elements which come from the elements of each $X_n$ (and it can be deduced from the universal property that every element of $X$ comes from some $X_n$, at least in the category of sets). On the other hand, a map $Xto X_n$ doesn't tell you any specific elements of $X$. So, you can't describe what an element of $X$ can look like using just a single $X_n$; you really need the whole sequence before you can even name a single element of $X$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      I object to the premise of your question, namely that the existence of a category with colimits but not limits has anything to do with the intuition you discuss. Note that in an arbitrary category, morphisms need not have any relation to functions. Moreover, given any category $C$, you can form the opposite category $C^{op}$ which reverses the directions of all the arrows. So, domains and codomains are really completely symmetric if you're just talking about arbitrary categories. For instance, if $C$ has all colimits but not all limits, then $C^{op}$ has all limits but not all colimits.



      The intuition that Ravi Vakil is talking about is not applicable to arbitrary categories. Instead, it's applicable to the sort of "concrete" categories we tend to encounter in practice (but not their opposite categories), where morphisms are closely related to functions and limits and filtered colimits are constructed similarly to how they are in the category of sets.



      As for the way you interpreted the intuition, I'm not entirely sure what you mean but it seems not entirely unreasonable. The way I might phrase it is that a map $X_nto X$ gives you actual elements of $X$ for each element of $X_0$. So, if $X$ is the colimit of the sequence, it has specific elements which come from the elements of each $X_n$ (and it can be deduced from the universal property that every element of $X$ comes from some $X_n$, at least in the category of sets). On the other hand, a map $Xto X_n$ doesn't tell you any specific elements of $X$. So, you can't describe what an element of $X$ can look like using just a single $X_n$; you really need the whole sequence before you can even name a single element of $X$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        I object to the premise of your question, namely that the existence of a category with colimits but not limits has anything to do with the intuition you discuss. Note that in an arbitrary category, morphisms need not have any relation to functions. Moreover, given any category $C$, you can form the opposite category $C^{op}$ which reverses the directions of all the arrows. So, domains and codomains are really completely symmetric if you're just talking about arbitrary categories. For instance, if $C$ has all colimits but not all limits, then $C^{op}$ has all limits but not all colimits.



        The intuition that Ravi Vakil is talking about is not applicable to arbitrary categories. Instead, it's applicable to the sort of "concrete" categories we tend to encounter in practice (but not their opposite categories), where morphisms are closely related to functions and limits and filtered colimits are constructed similarly to how they are in the category of sets.



        As for the way you interpreted the intuition, I'm not entirely sure what you mean but it seems not entirely unreasonable. The way I might phrase it is that a map $X_nto X$ gives you actual elements of $X$ for each element of $X_0$. So, if $X$ is the colimit of the sequence, it has specific elements which come from the elements of each $X_n$ (and it can be deduced from the universal property that every element of $X$ comes from some $X_n$, at least in the category of sets). On the other hand, a map $Xto X_n$ doesn't tell you any specific elements of $X$. So, you can't describe what an element of $X$ can look like using just a single $X_n$; you really need the whole sequence before you can even name a single element of $X$.






        share|cite|improve this answer









        $endgroup$



        I object to the premise of your question, namely that the existence of a category with colimits but not limits has anything to do with the intuition you discuss. Note that in an arbitrary category, morphisms need not have any relation to functions. Moreover, given any category $C$, you can form the opposite category $C^{op}$ which reverses the directions of all the arrows. So, domains and codomains are really completely symmetric if you're just talking about arbitrary categories. For instance, if $C$ has all colimits but not all limits, then $C^{op}$ has all limits but not all colimits.



        The intuition that Ravi Vakil is talking about is not applicable to arbitrary categories. Instead, it's applicable to the sort of "concrete" categories we tend to encounter in practice (but not their opposite categories), where morphisms are closely related to functions and limits and filtered colimits are constructed similarly to how they are in the category of sets.



        As for the way you interpreted the intuition, I'm not entirely sure what you mean but it seems not entirely unreasonable. The way I might phrase it is that a map $X_nto X$ gives you actual elements of $X$ for each element of $X_0$. So, if $X$ is the colimit of the sequence, it has specific elements which come from the elements of each $X_n$ (and it can be deduced from the universal property that every element of $X$ comes from some $X_n$, at least in the category of sets). On the other hand, a map $Xto X_n$ doesn't tell you any specific elements of $X$. So, you can't describe what an element of $X$ can look like using just a single $X_n$; you really need the whole sequence before you can even name a single element of $X$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        Eric WofseyEric Wofsey

        187k14215344




        187k14215344






























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