Fully faithful nerve ⟹ injective on objects?

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If $i: C ⟶ D$ (where $C$ is a small subcategory of $D$) is the inclusion functor, and $N_i := rm Hom(i(=), -)$ is the nerve of $i$, is it true that $N_i$ is injective on objects as soon as it is fully faithful?
(because if it is fully faithful, $i$ is dense, so I venture that this may be enough to deduce the injectivity of $N_i$ on objects)







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    If $i: C ⟶ D$ (where $C$ is a small subcategory of $D$) is the inclusion functor, and $N_i := rm Hom(i(=), -)$ is the nerve of $i$, is it true that $N_i$ is injective on objects as soon as it is fully faithful?
    (because if it is fully faithful, $i$ is dense, so I venture that this may be enough to deduce the injectivity of $N_i$ on objects)







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      If $i: C ⟶ D$ (where $C$ is a small subcategory of $D$) is the inclusion functor, and $N_i := rm Hom(i(=), -)$ is the nerve of $i$, is it true that $N_i$ is injective on objects as soon as it is fully faithful?
      (because if it is fully faithful, $i$ is dense, so I venture that this may be enough to deduce the injectivity of $N_i$ on objects)







      share|cite|improve this question












      If $i: C ⟶ D$ (where $C$ is a small subcategory of $D$) is the inclusion functor, and $N_i := rm Hom(i(=), -)$ is the nerve of $i$, is it true that $N_i$ is injective on objects as soon as it is fully faithful?
      (because if it is fully faithful, $i$ is dense, so I venture that this may be enough to deduce the injectivity of $N_i$ on objects)









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      asked Aug 26 at 8:50









      Cooke4

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          It's an odd question, because this depends on your conventions about homsets. Under many conventions, $N_i$ is injective on objects as soon as $C$ admits maps into every object of $D$, the point being that $Hom(c,d)$ is never equal to $Hom(c',d')$, unless perhaps they're both empty. Many foundations assume this disjointness of homsets. But it's awkward to think about equality of sets, and it's hard to see what this would do for you. In any case, this has nothing to do with density of $i$, and you could make your conventions so that $N_i$ needn't be injective on objects even with $i$ an identity if you really wanted: just take a category with two isomorphic objects such that all four morphisms are equal.






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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            up vote
            2
            down vote



            accepted










            It's an odd question, because this depends on your conventions about homsets. Under many conventions, $N_i$ is injective on objects as soon as $C$ admits maps into every object of $D$, the point being that $Hom(c,d)$ is never equal to $Hom(c',d')$, unless perhaps they're both empty. Many foundations assume this disjointness of homsets. But it's awkward to think about equality of sets, and it's hard to see what this would do for you. In any case, this has nothing to do with density of $i$, and you could make your conventions so that $N_i$ needn't be injective on objects even with $i$ an identity if you really wanted: just take a category with two isomorphic objects such that all four morphisms are equal.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              It's an odd question, because this depends on your conventions about homsets. Under many conventions, $N_i$ is injective on objects as soon as $C$ admits maps into every object of $D$, the point being that $Hom(c,d)$ is never equal to $Hom(c',d')$, unless perhaps they're both empty. Many foundations assume this disjointness of homsets. But it's awkward to think about equality of sets, and it's hard to see what this would do for you. In any case, this has nothing to do with density of $i$, and you could make your conventions so that $N_i$ needn't be injective on objects even with $i$ an identity if you really wanted: just take a category with two isomorphic objects such that all four morphisms are equal.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                It's an odd question, because this depends on your conventions about homsets. Under many conventions, $N_i$ is injective on objects as soon as $C$ admits maps into every object of $D$, the point being that $Hom(c,d)$ is never equal to $Hom(c',d')$, unless perhaps they're both empty. Many foundations assume this disjointness of homsets. But it's awkward to think about equality of sets, and it's hard to see what this would do for you. In any case, this has nothing to do with density of $i$, and you could make your conventions so that $N_i$ needn't be injective on objects even with $i$ an identity if you really wanted: just take a category with two isomorphic objects such that all four morphisms are equal.






                share|cite|improve this answer












                It's an odd question, because this depends on your conventions about homsets. Under many conventions, $N_i$ is injective on objects as soon as $C$ admits maps into every object of $D$, the point being that $Hom(c,d)$ is never equal to $Hom(c',d')$, unless perhaps they're both empty. Many foundations assume this disjointness of homsets. But it's awkward to think about equality of sets, and it's hard to see what this would do for you. In any case, this has nothing to do with density of $i$, and you could make your conventions so that $N_i$ needn't be injective on objects even with $i$ an identity if you really wanted: just take a category with two isomorphic objects such that all four morphisms are equal.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Aug 26 at 17:09









                Kevin Carlson

                29.5k23066




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