The following result appears in Charles Weibel's book An introduction to homological algebra as lemma 2.6.14 in page 56.
Lemma Let $I$ be a filtered category and let $A:I\rightarrow R\text{-mod}$ be a functor. Then for every $I$, the kernel of the canonical map $\lambda_i:A_i\rightarrow \text{colim}_{\rightarrow}(A_i)$ is union of kernel of maps $A_i\rightarrow A_j$.
I do not understand his proof. He first claims we can explicitly write what the colimit is, namely the direct sum of all $A_i$, i.e. $\oplus A_i$. Is it true? I think that is certainly wrong.
Second question is if we explicitly describe the colimit as $\oplus A_i$, would the kernel of the map $A_i\rightarrow \oplus A_i$ become $0$ then?
Third question: Weibel claims in his book that if $\lambda_i(a_i)=0$ then there exists $\varphi_{jk}:j\rightarrow k$ such that $\lambda_i(a_i)=\Sigma \lambda_k(\varphi_{jk}(a_k))-\lambda_j(a_j)$. Wouldn't that be trivial? Since on the left hand side it is zero and on the right hand side for each summand it is zero by the definition of colimit. Plus if possible can someone explain to me what Weibel is doing after achieving the equation $\lambda_i(a_i)=\Sigma \lambda_k(\varphi_{jk}(a_k))-\lambda_j(a_j)$?
My questions might be stupid and thank you in advance.
I see Ravenel has quite a sloppy scan version of Weibel's book (some lettes are misplaced or missing...)
Well, Weibel does not say that the colimit is a direct sum of the $A_i$, but instead that every element in the colimit can be written as a finite sum of some sort; using the assumption that your category of indices is filtered, now you know that any such finite family of indices maps in one of the summands, and that's where you define the $R$-module operations; the quotient that defines the filtered colimit gives that this is a well-defined structure of $R$-module on the filtered colimit of the underlying sets of the $R$-modules in your diagram.