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. $\square$

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