Why is local cohomology $R$-linear functor? Let $ R = k[x_1, \ldots, x_n] $ and $\mathfrak{m} = (x_1, \ldots, x_n)$. Let $N$ be an $R$-module, then the $\mathfrak{m}$-torsion functor is defined as $\Gamma_{\mathfrak{m}}(M) = \{ m \in M \text{ s.t.} \mathfrak{m}^nr = 0 \text{ for some } n \in \mathbb{N} \}$.
Now, in Brodmann and Sharps Local cohomology 1.2.1, it says:

Since $\Gamma_{\mathfrak{m}}(-)$ is covariant and $R$-linear, it is automatic that each local cohomology functor $H_{\mathfrak{m}}^{i}(-), i \geq 0$ is covariant and $R$-linear.

Why is this true?
It is clear for me, why $\Gamma_{\mathfrak{m}}(-)$ is a $R$-linear, left-exact functor.
I also know that for a short exact sequence of $R$-modules $$0 \to A \overset{f} \to B \to C \to 0,$$ there is a long exact sequence of cohomology modules s.t. we get $$H_{\mathfrak{m}}^{i}(A) \overset{H_{\mathfrak{m}}^{i}(f)} \to H_{\mathfrak{m}}^{i}(B) \to H_{\mathfrak{m}}^{i}(C).$$
So it makes sense that $H_{\mathfrak{m}}^{i}(-)$ is a functor for all $i \geq 0$. But why does it have to be $R$-linear?
While the map $f$ is from $A$ to $B$, the map $H_{\mathfrak{m}}^{i}(f)$ is from $\ker(\Gamma_{\mathfrak{m}}(d_A^{i}))/Im(\Gamma_{\mathfrak{m}}(d_A^{i-1}))$ to $\ker(\Gamma_{\mathfrak{m}}(d_B^{i}))/Im(\Gamma_{\mathfrak{m}}(d_B^{i-1}))$, where $d_A^{i}$ are the maps in an injective resolution of $A$ and $d_B^{i}$ in an injective resolution of $B$ respectively. Since the choice of injective resolution of $A$ and $B$ doesn't matter, how do I know so much about the map $H_{\mathfrak{m}}^{i}(f)$?
Clearly, $ H_{\mathfrak{m}}^{i}(f)$ depends on $f$, but how is it defined, resp. how does one see that is $R$-linear?
(My goal is to show that if $f$ is multiplication by $x \in R$, then $H_{\mathfrak{m}}^{i}(f)$ is still multiplication by $x \in R$.)
 A: First of all, "$H^i_{\mathfrak{m}}$ is $R$-linear" does not mean "$H^i_{\mathfrak{m}}(f)$ is $R$-linear for all $f$". Instead, it means $$H^i_{\mathfrak{m}}(\alpha f + g) = \alpha H^i_{\mathfrak{m}}(f) + H^i_{\mathfrak{m}}(g)$$ for all $f,g : A \to B$ and all $\alpha \in R$. However, the statement "$H^i_{\mathfrak{m}}(f)$ is $R$-linear for all $f$" is also true (and necessary in order for us to interpret the RHS of the above equation)! Once we unravel the definition of $H^i_{\mathfrak{m}}(f)$, this will be easy to prove.
So, it's not exactly true that the choice of injective resolution doesn't matter. Different injective resolutions for $A$ may actually yield different sets for $H^i_{\mathfrak{m}}(A)$. However, all the possible versions of $H^i_{\mathfrak{m}}(A)$ are isomorphic, and these isomorphisms are compatible with the maps $H^i_{\mathfrak{m}}(f)$. As a result of all this, we can compute cohomology groups (and maps between them) using any injective resolution we want (and just remember that the result is only well-defined up to isomorphism).
Well, how is $H^i_{\mathfrak{m}}(f)$ defined? First, we choose injective resolutions for $A$ and $B$:
$\require{AMScd}$
\begin{CD}
0 @>>> A @>>> A_0 @>>> A_1 @>>> \cdots\\
@. @VfVV \\
0 @>>> B @>>> B_0 @>>> B_1 @>>> \cdots
\end{CD}
Now, we fill this in to a morphism of complexes:
\begin{CD}
0 @>>> A @>>> A_0 @>d^0_A>> A_1 @>d^1_A>> \cdots\\
@. @VfVV @Vf_0VV @Vf_1VV \\
0 @>>> B @>>> B_0 @>d^0_B>> B_1 @>d^1_B>> \cdots
\end{CD}
Important exercise: how do we know this can be done? Hint: build the maps inductively, and use the definition of "injective module".
Next, we apply $\Gamma_{\mathfrak{m}}$ to the deleted resolutions:
\begin{CD}
0 @>>> \Gamma_{\mathfrak{m}}(A_0) @>\Gamma_{\mathfrak{m}}(d^0_A)>> \Gamma_{\mathfrak{m}}(A_1) @>\Gamma_{\mathfrak{m}}(d^1_A)>> \cdots\\
@. @V\Gamma_{\mathfrak{m}}(f_0)VV @V\Gamma_{\mathfrak{m}}(f_1)VV \\
0 @>>> \Gamma_{\mathfrak{m}}(B_0) @>\Gamma_{\mathfrak{m}}(d^0_B)>> \Gamma_{\mathfrak{m}}(B_1) @>\Gamma_{\mathfrak{m}}(d^1_B)>> \cdots
\end{CD}
Finally, we can state the definition of $H^i_{\mathfrak{m}}(f)$:
$$H^i_{\mathfrak{m}}(f)([x]) = [\Gamma_{\mathfrak{m}}(f_i)(x)]$$
for $x \in \ker(\Gamma_{\mathfrak{m}}(d^i_A))$, where $[\bullet]$ means "the equivalence class (aka coset) of $\bullet$".
Important exercise: This is a well-defined function, i.e. $[\Gamma_{\mathfrak{m}}(f_i)(x)] = [\Gamma_{\mathfrak{m}}(f_i)(y)]$ whenever $[x] = [y]$ for $x, y \in \ker(\Gamma_{\mathfrak{m}}(d^i_A))$.
Important note: This construction of $H^i_{\mathfrak{m}}(f)$ relied on three arbitrary choices:

*

*An injective resolution of $A$

*An injective resolution of $B$

*An extension of $f$ to a morphism of complexes

We've already discussed how the choices of injective resolution for $A$ and $B$ don't end up mattering, up to isomorphism. Moreover, it turns out that choosing a different morphism of complexes would result in exactly the same map $H^i_{\mathfrak{m}}(f)$ (not just up to isomorphism)! Thus, $H^i_{\mathfrak{m}}(f)$ is well-defined.

Once we have this definition, we can check that $H^i_{\mathfrak{m}}(f)$ is always $R$-linear! We let $a, a' \in H^i_{\mathfrak{m}}(A)$ and $\lambda \in R$ be arbitrary. Then $a = [x], a' = [x']$ for some $x, x' \in \ker(\Gamma_{\mathfrak{m}}(d^i_A))$. Now
\begin{align*}
H^i_{\mathfrak{m}}(f)(\lambda a + a') &= H^i_{\mathfrak{m}}(f)(\lambda [x] + [x'])\\
&= H^i_{\mathfrak{m}}(f)([\lambda x + x']) & \text{(definition of quotient module)}\\
&= [\Gamma_{\mathfrak{m}}(f_i)(\lambda x + x')] & \text{(definition of } H^i_{\mathfrak{m}}(f)\text{)}\\
&= [\lambda \Gamma_{\mathfrak{m}}(f_i)(x) + \Gamma_{\mathfrak{m}}(f_i)(x')] & \text{(} \Gamma_{\mathfrak{m}}(f_i) \text{ is } R\text{-linear)}\\
& = \lambda [\Gamma_{\mathfrak{m}}(f_i)(x)] + [\Gamma_{\mathfrak{m}}(f_i)(x')] & \text{(definition of quotient module)}\\
&= \lambda H^i_{\mathfrak{m}}(f)([x]) + H^i_{\mathfrak{m}}(f)([x']) & \text{(definition of } H^i_{\mathfrak{m}}(f)\text{)}\\
&= \lambda H^i_{\mathfrak{m}}(f)(a) + H^i_{\mathfrak{m}}(f)(a')
\end{align*}
and we're done!

I want to emphasize that this is where the definition of $H^i_{\mathfrak{m}}$ as a functor comes from (not the long exact sequence)! Indeed, the map $H^i_{\mathfrak{m}}(f)$ in the long exact sequence doesn't make sense to write down before we've defined $H^i_{\mathfrak{m}}(f)$ (i.e. before we've made $H^i_{\mathfrak{m}}$ into a functor).
We still haven't shown that $H^i_{\mathfrak{m}}$ is an $R$-linear functor, but I believe you can do that from here! Definitely begin by proving that $\Gamma_m$ is $R$-linear. Let me know if you get stuck!
