Are all the kernel of a r-module isomorphic?

Question

I'm fresh to homological algebra, and our textbook is Weibel's.

When I was showing the Exercise 1.2.2 of Weibel's book, I'm in the following "path" so that the notations of kernel, monic and monomorphism are the same

(1) Every kernel is monic;

(2) Every monic morphism in $R$-$\mathbf{Mod}$ is a monomorphism;

(3) Every monomorphism in $R$-$\mathbf{Mod}$ is a kernel of a morphism.

When I'm doing (3) the monomorphism $f:A\to B$, I first define the quotient map $\pi: B\to B/\mathrm{im}{f}$ and show that the inclusion $i:\mathrm{im}{f}\to B$ is a kernel of $\pi$, and then use the universal property of $i$ to show the universal property of $f:A\to B$. And in my argument, I use the monomorphism property to show that there is a bijection from $A$ to $\mathrm{im}f$, and thus the domain of $f$ and $i$ are "the same" in some sense, and their codomain are both $\mathrm{im}f$, and it seems that this "the same" preserves the composition of morphisms.

Here is my question, are all kernels of a morphism the same in some sense?

I've read that the kernel and the usual $\ker{f}=\{a\in A: f(a)=0\}$ are the same, and the latter one is unique obviously, then all the kernel of a map in $R$-$\mathbf{Mod}$ would be the same in some sense. But in what sense?

Answer 1

The categorical definition of kernel is a special case of limit, and limits are unique up to isomorphism.

Now this means that, if $k_1:K_1\to A$ and $k_2:K_2\to A$ both satisfy the universal property of being the kernel of a (nonnecessarily injective) morphism $f:A\to B$, then there is a (unique) isomorphism $u:K_1\to K_2$ such that $k_1=k_2\circ u$.

We have $f\circ k_i=0$ and thus by the kernel property of $k_j$, there's a unique morphism $u_{ij}:K_i\to K_j$ making $k_i=k_j\circ u_{ij}$ for $i,j=1,2$.
For $i=j$, observe that $1_{K_i}$ and $u_{i'i}\circ u_{ii'}$ both satisfy the above equation, so by uniqueness, $u_{ii}=1_{K_i}=u_{i'i}\circ u_{ii'}$.

On the other hand, if $k_2$ is a kernel of $f$, and $u:K_1\to K_2$ is an isomorphism, then it is easy to see that $k_2\circ u$ is again a kernel of $f$.