# Monday, February 15

## 2.5: Right-Derived Functors

:::{.remark}
Today: right-derived functors of a left-exact functor.
Luckily we can use some opposite category tricks which save us some work of re deriving everything.
:::

:::{.definition title="Right Derived Functors"}
Let \( F: \cat{A} \to \cat{B}  \) be left-exact where \( \cat{A} \) has enough injectives.
Given \( A \in \cat{A} \), fix an injective resolution \( 0 \to A \mapsvia{\eps} I\) and define
\[
R^i \cat{F} \da H^i( FA ) && i \geq 0 
.\]
:::

:::{.remark}
Then 
\[
0 \to FA \mapsvia{F\eps} FI^0 \mapsvia{Fd^0} FI^1
\]
is exact,
and 
\[
R^0 FA = \ker F(d^0) / \gens{ 0 } = \im F\eps \cong FA
,\]
and so there is naturally an isomorphism $R^0 F \cong F$.
Observe that \( F \) yields a right-exact functor $F\op: \cat{A}\op \to \cat{B}\op$, where we note that $F\op (f\op) = F(f)\op$.
Note that taking the opposite category sends injectives to projectives and so \( \cat{A}\op \) has enough projectives.
This means that $L_i F\op$ are defined using the projective resolution $I$, so we have
\[
R^i F(A) = (L_i F\op)\op
.\]
Thus all results about left-derived functors translate to right-derived functors:

- $R_i F$ is independent of the choice of injective resolution, up to a natural isomorphism.
- If $A$ is injective, then $R^{i>0} F(A) = 0$.
- The collection \( \ts{ R^i F } _{i\geq 0 } \) forms a universal cohomological \( \delta\dash \)functor for \( F \).
- An object \( Q \in \cat{A} \) is **$F\dash$acyclic** if $R^{>0}F(Q) = 0$.
- $R^iF$ can be computed using $F\dash$acyclic objects instead of injective resolutions.
:::

:::{.definition title="?"}
Fix a right \(R\dash\)module \( M \in \modr \), then $F \da \Hom_{\modr}(M, \wait): \modr \to \Ab$ is a left-exact functor.
Its right-derived functors are **ext functors** and denoted $\Ext_{\modr}^i(M, \wait)$.
:::

:::{.example title="?"}
\[ \Ext_{\modr}^i(M, A) = (R^i F)(A) = [ R^i \Hom_{\modr}(M, \wait) ] (A) .\]
:::

:::{.remark}
Exercises 2.5.1, 2.5.2 are important extensions of our existing characterizations of injectives and projectives in $\modr$.
These upgrade the characterization involving $\Hom$ to one involving $\Ext$.
[^note_typo]

[^note_typo]: 
Note the typo in 2.5.1.3, it should say the following:
"$B$ is $\Hom_{R}(A, \wait)$ is acyclic for all $A$."

:::

:::{.remark}
Fix $B\in \modr$ and consider $G\da \Hom_{\modr}(\wait, B): \modr \to \Ab$.
Then $G$ is still left-exact, but is now *contravariant*.
We can regard it as a covariant functor left-exact functor $G: \modr \op \to \Ab$.
So we define $R^i G(A)$ by an injective resolution of $A$ in \( \cat{A}\op \), and this is the same as a projective resolution of $A$ in \( \cat{A} \).
So apply \( G \) and take cohomology.
It turns out that 
\[ 
R^i \Hom_{\modr}(\wait, B) \cong R^i \Hom_{\modr}(A, \wait)(B) \da \Ext^i_{\rmod}(A, B) 
,\] 
so we can use the same notation $\Ext^i_R(\wait, B)$ for both cases.
:::

## 2.6: Adjoint Functors and Left/Right Exactness

:::{.slogan}
$\wait$ adjoints are $\wait\op$ exact, since $\wait$ adjoints have $\wait\dash$derived functors.
:::

:::{.theorem title="Exactness of adjoint functors"}
Let 
\[
\adjunction{L}{R}{ \cat{A} } { \cat{B} }
\]
be an adjoint pair of functors.
Then there exists a natural isomorphism
\[
\tau_{AB}: \Hom_{\cat{B}}(LA, B) \mapsvia{\sim} \Hom_{\cat{A}}(A, RB) \quad  \forall A\in \cat{A}, B\in \cat{B}
.\]
Moreover, 

- $L$ is right exact, and
- $B$ is left exact.

:::

:::{.proposition title="1.6: Yoneda"}
A sequence
\[
A \mapsvia{\alpha} B \mapsvia{\beta} C
\]
is exact in \( \cat{A} \) if and only if for all \( M \in \Ob( \cat{A} ) \), the sequence
\[
\Hom_{\cat{A}} (M, A)
\mapsvia{\alpha^* \da \alpha\circ \wait} 
\Hom_{\cat{A}} (M, B)
\mapsvia{\beta^* \da \beta \circ \wait} 
\Hom_{\cat{A}} (M, C)
\]
is exact.
:::

:::{.proof title="?"}
\envlist

1. Take $M=A$, then $0 = \beta^* \alpha^*(\one_A) = \beta \alpha \one = \beta \alpha$.
  Thus $\im \alpha \subseteq \ker \beta$.

2. Take $M = \ker \beta$ and consider the inclusion \( \iota: \ker M \injects B \), then \( \beta^*(\iota) = \beta \iota = 0 \) and thus \( \iota \in \ker \beta^* = \im \alpha^* \).
  So there exists \( \sigma\in \Hom( \ker \beta, A) \) such that \( \iota = \alpha^* (\sigma) \da \alpha \sigma \), and thus \( \ker \beta = \im \iota \subset \im \alpha \).

Thus $\ker \beta= \im \alpha$, yielding exactness of the bottom sequence.
:::

:::{.proof title="of theorem"}
We'll first prove that $R$ is left-exact.
Take a SES in $B$, say
\[
0 \to B' \to B \to B'' \to 0
.\]
Apply the left-exact covariant functor $\Hom_{\cat{B}}(LA, \wait)$ followed by $\tau$:

\begin{tikzcd}
	0 && { \Hom_{\cat{B}} (LA, B') } && { \Hom_{\cat{B}} (LA, B) } && { \Hom_{\cat{B}} (LA, B'')} \\
	\\
	0 && {\Hom_{\cat{B}} (A, RB')} && {\Hom_{\cat{B} }(A, RB)} && {\Hom_{\cat{B}} (A, RB'')}
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow["{\tau_{AB}}"{description}, squiggly, from=1-3, to=3-3]
	\arrow["{\tau_{AB}}"{description}, squiggly, from=1-5, to=3-5]
	\arrow["{\tau_{AB}}"{description}, squiggly, from=1-7, to=3-7]
	\arrow[dashed, from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

The bottom sequence is exact by naturality of $\tau$.
Now applying the Yoneda lemma, we obtain an exact sequence
\[
0 \to 
\Hom_{\cat{A}}(A, RB')
\to
\Hom_{\cat{A}}(A, RB)
\to
\Hom_{\cat{A}}(A, RB'')
.\]

So $R$ is left exact.
Now $L\op: \cat{A} \to \cat{B}$ is right adjoint to $R\op$, so $L\op$ is left exact and thus $L$ is right exact.
:::

## Tensor Product Functors and Tor

:::{.remark}
Let 

- $R, S \in \Ring$, 
- $B\in \bimod{R}{S}$, 
- $C\in \mods{S}$.

Then $\Hom_{S}(B, C)\in \modr$ in a natural way: given $f:B\to C$, define $(f\cdot r)(b) = f(rb)$.
:::

:::{.exercise title="?"}
Check that this is a well-defined morphism of right \(S\dash\)modules.
:::

:::{.remark}
We saw this structure earlier with $S=\ZZ$, see p.41.
:::

:::{.proposition title="Tensor-Hom adjunction"}
Fix $R,S$ and ${}_R B_S$ as above.
Then for every \( A \in \modr \) and \( C\in \mods\dash S \) there is a natural isomorphism
\[
\tau: \Hom_S( A\tensor_R B, C ) &\mapsvia{\sim} \Hom_R(A, \Hom_S(B, C) ) \\
f &\mapsto g(a)(b) = f(a\tensor b) \\
f(a\tensor b) = g(a)(b) &\mapsfrom g
.\]

Note that the tensor product is a right $S\dash$module, and the hom on the right is a right \(R\dash\)module, so these expressions make sense.
Here $B$ is fixed, so $A$ and $C$ are variables and this is a statement about bifunctors
\[
\wait \tensor_R B: \modr \to \modsright{S} 
,\]
which is left adjoint to 
\[
\Hom_S(B, \wait): \modsright{S} \to \modr
.\]
So the former is a left adjoint and the latter is a right adjoint, so by the theorem, $\wait \tensor_R B$ is right exact.
:::

:::{.remark}
If $B$ is only a left \(R\dash\)module, we can always take $S = \ZZ$, which makes this into a functor
\[
\wait \tensor_R B: \modr \to \Ab
.\]
Since this is a right exact functor from a category with enough injectives, we can define left-derived functors.
:::

:::{.definition title="?"}
Let $B\in \bimod{R}{S}$ and let 
\[
T(\wait) \da \wait\tensor_R B: \modr \to \modsright{S}
.\]
Then define $\Tor_n^R(A, B) \da L_n T(A)$.
:::

:::{.remark}
Note that these are easier to work with, since they're covariant in both variables.
:::