Lemma 13.14.3. Assumptions and notation as in Situation 13.14.1. Let $f : X \to Y$ be a morphism of $\mathcal{D}$.

1. If $RF$ is defined at $X$ and $Y$ then there exists a unique morphism $RF(f) : RF(X) \to RF(Y)$ between the values such that for any commutative diagram

$\xymatrix{ X \ar[d]_ f \ar[r]_ s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$

with $s, s' \in S$ the diagram

$\xymatrix{ F(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & RF(X) \ar[d] \\ F(Y) \ar[r] & F(Y') \ar[r] & RF(Y) }$

commutes.

2. If $LF$ is defined at $X$ and $Y$ then there exists a unique morphism $LF(f) : LF(X) \to LF(Y)$ between the values such that for any commutative diagram

$\xymatrix{ X' \ar[d]_{f'} \ar[r]_ s & X \ar[d]^ f \\ Y' \ar[r]^{s'} & Y }$

with $s, s'$ in $S$ the diagram

$\xymatrix{ LF(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & F(X) \ar[d] \\ LF(Y) \ar[r] & F(Y') \ar[r] & F(Y) }$

commutes.

Proof. Part (1) holds if we only assume that the colimits

$RF(X) = \mathop{\mathrm{colim}}\nolimits _{s : X \to X'} F(X') \quad \text{and}\quad RF(Y) = \mathop{\mathrm{colim}}\nolimits _{s' : Y \to Y'} F(Y')$

exist. Namely, to give a morphism $RF(X) \to RF(Y)$ between the colimits is the same thing as giving for each $s : X \to X'$ in $\mathop{\mathrm{Ob}}\nolimits (X/S)$ a morphism $F(X') \to RF(Y)$ compatible with morphisms in the category $X/S$. To get the morphism we choose a commutative diagram

$\xymatrix{ X \ar[d]_ f \ar[r]_ s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$

with $s, s'$ in $S$ as is possible by MS2 and we set $F(X') \to RF(Y)$ equal to the composition $F(X') \to F(Y') \to RF(Y)$. To see that this is independent of the choice of the diagram above use MS3. Details omitted. The proof of (2) is dual. $\square$

Comment #8375 by on

To hive more hints at the reader, maybe one could expand "to see that this is independent of the choice of the diagram above use MS3 and the fact that $Y/S$ is filtered. To see after that these well-defined morphisms $F(X')\to RF(Y)$ are compatible with the morphisms in $X/S$, use also MS3 plus filteredness of $Y/S$. Details ommited."

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