Kernel of a $R$-linear map is uniquely determined? In my algebra textbook there is the following exercise:
Let $M$ and $N$ be two left $R$-modules and $f:M\longrightarrow N$ a $R$-linear map. Show the kernel $L$ of $f$ is uniquely determined by the following universal property:
$(i)$ There is a unique $R$-linear map $\ell:L\longrightarrow M$ such that $f\ell=0$.
$(ii)$ If $\ell^\prime:L^\prime\longrightarrow M$ is a linear map such that $f\ell^\prime=0$ there is a unique $R$-linear map $u:L^\prime\longrightarrow L$ such that $\ell^\prime=\ell u$.
The main problem is that I didn't understand what I'm supposed to do, so I don't even know where to start.
Can anyone help me out?
Thanks
 A: The issue here is that you need to know what you mean by "unique". You can show that such an object $L$ exists ; just take it to be the submodule of $M$ of elements mapped to $0$ under $f$, and check the two conditions. For "unicity", if you have two modules $L_1,L_2$ that satisfy property (i) and (ii), you cannot hope to show that $L_1 = L_2$, but you can certainly show that $L_1 \simeq L_2$ as $R$-modules, and you can do that assuming that (i) and (ii) are satisfied by $L_1$ and $L_2$. We usually say that $L$ is "unique up to unique isomorphism". 
In category theory, the two properties (i) and (ii) define an object called "a kernel of $f$", where $f$ is some arrow in the category. Any two kernels are then isomorphic. For category theorists, this is better because it removes the "explicit construction" of the picture (i.e. your representation of the kernel using the submodule of elements mapped to $0$ by $f$) and only preserves the "important properties" of the kernel.
Feel free to ask if you want more hints.
Hope that helps,
