Lemma 10.8.7. Let $(M_ i, \mu _{ij})$, $(N_ i, \nu _{ij})$ be systems of $R$-modules over the same preordered set. A morphism of systems $\Phi = (\phi _ i)$ from $(M_ i, \mu _{ij})$ to $(N_ i, \nu _{ij})$ induces a unique homomorphism

$\mathop{\mathrm{colim}}\nolimits \phi _ i : \mathop{\mathrm{colim}}\nolimits M_ i \longrightarrow \mathop{\mathrm{colim}}\nolimits N_ i$

such that

$\xymatrix{ M_ i \ar[r] \ar[d]_{\phi _ i} & \mathop{\mathrm{colim}}\nolimits M_ i \ar[d]^{\mathop{\mathrm{colim}}\nolimits \phi _ i} \\ N_ i \ar[r] & \mathop{\mathrm{colim}}\nolimits N_ i }$

commutes for all $i \in I$.

Proof. Write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ and $N = \mathop{\mathrm{colim}}\nolimits N_ i$ and $\phi = \mathop{\mathrm{colim}}\nolimits \phi _ i$ (as yet to be constructed). We will use the explicit description of $M$ and $N$ in Lemma 10.8.2 without further mention. The condition of the lemma is equivalent to the condition that

$\xymatrix{ \bigoplus _{i\in I} M_ i \ar[r] \ar[d]_{\bigoplus \phi _ i} & M \ar[d]^\phi \\ \bigoplus _{i\in I} N_ i \ar[r] & N }$

commutes. Hence it is clear that if $\phi$ exists, then it is unique. To see that $\phi$ exists, it suffices to show that the kernel of the upper horizontal arrow is mapped by $\bigoplus \phi _ i$ to the kernel of the lower horizontal arrow. To see this, let $j \leq k$ and $x_ j \in M_ j$. Then

$(\bigoplus \phi _ i)(x_ j - \mu _{jk}(x_ j)) = \phi _ j(x_ j) - \phi _ k(\mu _{jk}(x_ j)) = \phi _ j(x_ j) - \nu _{jk}(\phi _ j(x_ j))$

which is in the kernel of the lower horizontal arrow as required. $\square$

Comment #1834 by Patrizio on

Would it not be easier just to use the universal property? In fact, We can take the composition M_i \rightarrow N_i \rightarrow lim N_i and then we know that there exists a unique morphism which factorizes by lim M_i, just by definition. Is that idea wrong?

Comment #1871 by on

Yes, I agree that works and I agree that it is better. If you feel so inclined, please edit the latex file and send it to me. Thanks, Johan

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