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$\underline {Background}$: 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.

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  • $\begingroup$ 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
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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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