On morphisms in an abelian category












0












$begingroup$


$underline {Background}$: Suppose ,we are in an abelian category $mathcal C$ and let $B in mathcal C$ be an object.



Let,$f_1,f_2 in Hom(B,B)$.



Since $mathcal C$ is an abelian category $f_1-f_2 in Hom(B,B)$.



$underline {Question(1)}$:what is the image of $f_1-f_2$?



i.e can we calculate this image in general in terms of kernel ,cokernel ,image of the morphisms $f_1$,$f_2$?



$underline { Guess}$: may be under some suitable condition it could be $Im(f_1)/Im(f_2)$ (the only natural thing that comes to my mind is this even though I do not see why Im$f_2$ should be embedded in Im$f_1$).But I do not see any clue how to proceed further.



$underline {Question (2)}$: even if we cannot represent it generally in terms of kernel,cokernel and image of $f_1,f_2 $ how to think of this image category theoretically,i.e diagrammatically.



Any help from anyone is welcome.






category-theory abelian-categories







share|cite|improve this question









$endgroup$












  • $begingroup$
    Why exactly do you think there should be something like this in Question 1? I mean, if I encounter such a question, the first thing I would do is to see if this holds for abelian groups or vector spaces over some field.
    $endgroup$
    – Paul K
    Dec 16 '18 at 16:47
















0












$begingroup$


$underline {Background}$: Suppose ,we are in an abelian category $mathcal C$ and let $B in mathcal C$ be an object.



Let,$f_1,f_2 in Hom(B,B)$.



Since $mathcal C$ is an abelian category $f_1-f_2 in Hom(B,B)$.



$underline {Question(1)}$:what is the image of $f_1-f_2$?



i.e can we calculate this image in general in terms of kernel ,cokernel ,image of the morphisms $f_1$,$f_2$?



$underline { Guess}$: may be under some suitable condition it could be $Im(f_1)/Im(f_2)$ (the only natural thing that comes to my mind is this even though I do not see why Im$f_2$ should be embedded in Im$f_1$).But I do not see any clue how to proceed further.



$underline {Question (2)}$: even if we cannot represent it generally in terms of kernel,cokernel and image of $f_1,f_2 $ how to think of this image category theoretically,i.e diagrammatically.



Any help from anyone is welcome.






category-theory abelian-categories







share|cite|improve this question









$endgroup$












  • $begingroup$
    Why exactly do you think there should be something like this in Question 1? I mean, if I encounter such a question, the first thing I would do is to see if this holds for abelian groups or vector spaces over some field.
    $endgroup$
    – Paul K
    Dec 16 '18 at 16:47














0












0








0





$begingroup$


$underline {Background}$: Suppose ,we are in an abelian category $mathcal C$ and let $B in mathcal C$ be an object.



Let,$f_1,f_2 in Hom(B,B)$.



Since $mathcal C$ is an abelian category $f_1-f_2 in Hom(B,B)$.



$underline {Question(1)}$:what is the image of $f_1-f_2$?



i.e can we calculate this image in general in terms of kernel ,cokernel ,image of the morphisms $f_1$,$f_2$?



$underline { Guess}$: may be under some suitable condition it could be $Im(f_1)/Im(f_2)$ (the only natural thing that comes to my mind is this even though I do not see why Im$f_2$ should be embedded in Im$f_1$).But I do not see any clue how to proceed further.



$underline {Question (2)}$: even if we cannot represent it generally in terms of kernel,cokernel and image of $f_1,f_2 $ how to think of this image category theoretically,i.e diagrammatically.



Any help from anyone is welcome.






category-theory abelian-categories







share|cite|improve this question









$endgroup$




$underline {Background}$: Suppose ,we are in an abelian category $mathcal C$ and let $B in mathcal C$ be an object.



Let,$f_1,f_2 in Hom(B,B)$.



Since $mathcal C$ is an abelian category $f_1-f_2 in Hom(B,B)$.



$underline {Question(1)}$:what is the image of $f_1-f_2$?



i.e can we calculate this image in general in terms of kernel ,cokernel ,image of the morphisms $f_1$,$f_2$?



$underline { Guess}$: may be under some suitable condition it could be $Im(f_1)/Im(f_2)$ (the only natural thing that comes to my mind is this even though I do not see why Im$f_2$ should be embedded in Im$f_1$).But I do not see any clue how to proceed further.



$underline {Question (2)}$: even if we cannot represent it generally in terms of kernel,cokernel and image of $f_1,f_2 $ how to think of this image category theoretically,i.e diagrammatically.



Any help from anyone is welcome.






category-theory abelian-categories




category-theory abelian-categories






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share|cite|improve this question











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asked Dec 16 '18 at 8:25









HARRYHARRY

889




889












  • $begingroup$
    Why exactly do you think there should be something like this in Question 1? I mean, if I encounter such a question, the first thing I would do is to see if this holds for abelian groups or vector spaces over some field.
    $endgroup$
    – Paul K
    Dec 16 '18 at 16:47


















  • $begingroup$
    Why exactly do you think there should be something like this in Question 1? I mean, if I encounter such a question, the first thing I would do is to see if this holds for abelian groups or vector spaces over some field.
    $endgroup$
    – Paul K
    Dec 16 '18 at 16:47
















$begingroup$
Why exactly do you think there should be something like this in Question 1? I mean, if I encounter such a question, the first thing I would do is to see if this holds for abelian groups or vector spaces over some field.
$endgroup$
– Paul K
Dec 16 '18 at 16:47




$begingroup$
Why exactly do you think there should be something like this in Question 1? I mean, if I encounter such a question, the first thing I would do is to see if this holds for abelian groups or vector spaces over some field.
$endgroup$
– Paul K
Dec 16 '18 at 16:47










1 Answer
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1












$begingroup$

Question 1: You cannot say anything about the image of $f_1-f_2$. Here is an example. Let $mathcal{C}$ be the category of real vector spaces and $B = mathbb{R}^n$. The set $text{End}(mathbb{R}^n)$ of endomorphisms $phi : mathbb{R}^n to mathbb{R}^n$ is a normed linear space and it is well-known that the set $text{GL}(mathbb{R}^n)$ of automorphisms is open in $text{End}(mathbb{R}^n)$.



Let $f_1 = id_B$ and $V subset mathbb{R}^n$ any linear subspace. Choose any linear map $g : mathbb{R}^n to mathbb{R}^n$ such that $text{im}(g) = V$. Hence for sufficiently small $epsilon > 0$ the linear map $f_2 = id_B + epsilon g$ belongs to $text{GL}(mathbb{R}^n)$.



Both $f_1, f_2$ are isomorphisms, thus their kernels and cokernels are trivial and their images are $mathbb{R}^n$. However, $f_1 - f_2 = - epsilon g$ whose image is $V$.



Question 2: I do not think that one can get a reasonable answer.






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    $begingroup$

    Question 1: You cannot say anything about the image of $f_1-f_2$. Here is an example. Let $mathcal{C}$ be the category of real vector spaces and $B = mathbb{R}^n$. The set $text{End}(mathbb{R}^n)$ of endomorphisms $phi : mathbb{R}^n to mathbb{R}^n$ is a normed linear space and it is well-known that the set $text{GL}(mathbb{R}^n)$ of automorphisms is open in $text{End}(mathbb{R}^n)$.



    Let $f_1 = id_B$ and $V subset mathbb{R}^n$ any linear subspace. Choose any linear map $g : mathbb{R}^n to mathbb{R}^n$ such that $text{im}(g) = V$. Hence for sufficiently small $epsilon > 0$ the linear map $f_2 = id_B + epsilon g$ belongs to $text{GL}(mathbb{R}^n)$.



    Both $f_1, f_2$ are isomorphisms, thus their kernels and cokernels are trivial and their images are $mathbb{R}^n$. However, $f_1 - f_2 = - epsilon g$ whose image is $V$.



    Question 2: I do not think that one can get a reasonable answer.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Question 1: You cannot say anything about the image of $f_1-f_2$. Here is an example. Let $mathcal{C}$ be the category of real vector spaces and $B = mathbb{R}^n$. The set $text{End}(mathbb{R}^n)$ of endomorphisms $phi : mathbb{R}^n to mathbb{R}^n$ is a normed linear space and it is well-known that the set $text{GL}(mathbb{R}^n)$ of automorphisms is open in $text{End}(mathbb{R}^n)$.



      Let $f_1 = id_B$ and $V subset mathbb{R}^n$ any linear subspace. Choose any linear map $g : mathbb{R}^n to mathbb{R}^n$ such that $text{im}(g) = V$. Hence for sufficiently small $epsilon > 0$ the linear map $f_2 = id_B + epsilon g$ belongs to $text{GL}(mathbb{R}^n)$.



      Both $f_1, f_2$ are isomorphisms, thus their kernels and cokernels are trivial and their images are $mathbb{R}^n$. However, $f_1 - f_2 = - epsilon g$ whose image is $V$.



      Question 2: I do not think that one can get a reasonable answer.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Question 1: You cannot say anything about the image of $f_1-f_2$. Here is an example. Let $mathcal{C}$ be the category of real vector spaces and $B = mathbb{R}^n$. The set $text{End}(mathbb{R}^n)$ of endomorphisms $phi : mathbb{R}^n to mathbb{R}^n$ is a normed linear space and it is well-known that the set $text{GL}(mathbb{R}^n)$ of automorphisms is open in $text{End}(mathbb{R}^n)$.



        Let $f_1 = id_B$ and $V subset mathbb{R}^n$ any linear subspace. Choose any linear map $g : mathbb{R}^n to mathbb{R}^n$ such that $text{im}(g) = V$. Hence for sufficiently small $epsilon > 0$ the linear map $f_2 = id_B + epsilon g$ belongs to $text{GL}(mathbb{R}^n)$.



        Both $f_1, f_2$ are isomorphisms, thus their kernels and cokernels are trivial and their images are $mathbb{R}^n$. However, $f_1 - f_2 = - epsilon g$ whose image is $V$.



        Question 2: I do not think that one can get a reasonable answer.






        share|cite|improve this answer









        $endgroup$



        Question 1: You cannot say anything about the image of $f_1-f_2$. Here is an example. Let $mathcal{C}$ be the category of real vector spaces and $B = mathbb{R}^n$. The set $text{End}(mathbb{R}^n)$ of endomorphisms $phi : mathbb{R}^n to mathbb{R}^n$ is a normed linear space and it is well-known that the set $text{GL}(mathbb{R}^n)$ of automorphisms is open in $text{End}(mathbb{R}^n)$.



        Let $f_1 = id_B$ and $V subset mathbb{R}^n$ any linear subspace. Choose any linear map $g : mathbb{R}^n to mathbb{R}^n$ such that $text{im}(g) = V$. Hence for sufficiently small $epsilon > 0$ the linear map $f_2 = id_B + epsilon g$ belongs to $text{GL}(mathbb{R}^n)$.



        Both $f_1, f_2$ are isomorphisms, thus their kernels and cokernels are trivial and their images are $mathbb{R}^n$. However, $f_1 - f_2 = - epsilon g$ whose image is $V$.



        Question 2: I do not think that one can get a reasonable answer.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 9:18









        Paul FrostPaul Frost

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