fix issue #20

Author: EricWay1024

ha/2-ab.typ

@@ -745,8 +745,10 @@ In other words, $P$ is projective if for any #sest $ses(X, Y, Z)$ in $cA$, $ ses
 
 #note[
   If $cA = SMod$ for some ring $S$, we have observed that $S$ (as an object of $SMod$) is a compact, projective generator. In this case, $R = end_S (S)$. We observe that any module homomorphism $phi: S -> S$ is uniquely determined by $phi(1) in S$ with $phi(s) = s phi(1)$, and the composition of two homomorphisms $phi_1 , phi_2 : S-> S$ is in the opposite direction of multiplication in $S$: $ phi_1 (phi_2(s)) = s phi_2(1) phi_1(1) $
-  Therefore, $R = end_S (S) = S^op$. Thus, indeed, we have $SMod$ is equivalent to $ModR$, which is $Mod$-$S^op$.
+  Therefore, $R = end_S (S) = S^op$. Thus, indeed, we have $SMod$ is equivalent to $ModR$, which is $ModS^op$.
 ]
+<opposite-ring-mod>
+
 
 // #remark[
 //   Using the definition of equivalence, you want to construct another functor in the opposite direction and show their composites are natural isomorphic to identity functors. Alternatively, you might also prove that the functor is fully faithful and essentially surjective, if you can.
@@ -819,7 +821,6 @@ In other words, $P$ is projective if for any #sest $ses(X, Y, Z)$ in $cA$, $ ses
 ]
 
 
-
 #theorem("Freyd-Mitchell Embedding Theorem")[
   If $cA$ is a small abelian category, there is a ring $R$ and an exact, fully faithful embedding functor $cA -> RMod$.
 ]

ha/4-enough.typ

@@ -338,33 +338,34 @@ With this proposition, we can prove that an abelian category has enough projecti
   If $I$ is an injective abelian group, then $hom_Ab (R, I)$ is an injective #rrm.
 ]
 #proof[
-  By @hom-module, $hom_Ab (R, I)$ is indeed a right $R$-module. Note that $hom_Ab (R, -)$ is right adjoint to $(- tpr R)$, which is simply the forgetful functor $ModR -> Ab$ and is thus exact. Therefore $hom_Ab (R, I)$ is injective in $RMod$.
+  By @hom-module, $hom_Ab (R, I)$ is indeed a right $R$-module. Note that $hom_Ab (R, -)$ is right adjoint to $(- tpr R)$, which is simply the forgetful functor $ModR -> Ab$ and is thus exact. Therefore by the previous proposition, $hom_Ab (R, I)$ is injective in $ModR$.
 ]
 
 #example[
-  $hom_Ab (R, QQ over ZZ)$ is injective.
+  $hom_Ab (R, QQ over ZZ)$ is injective in $ModR$.
 ]
 
 #proposition[
   $RMod$ has enough injectives.
 ]
 
 #proof[
+  Since $RMod$ is equivalent to $ModR^(op)$ (see @opposite-ring-mod), it suffices to prove that $ModR$ has enough injectives for any $R$. 
   Define map
   $
-    I : RMod &-> RMod, \
+    I : ModR &-> ModR, \
     M &|-> product_(homr(M, hom_Ab (R, QQ over ZZ))) hom_Ab (R, QQ over ZZ).
   $
 
-  For any left $R$-module $M$,
+  For any right $R$-module $M$,
   $I(M)$ is injective as a product of injectives, and there is a canonical morphism
   $
     e_M: M &-> I(M ), \
     m &|-> (phi(m))_(phi in homr(M, hom_Ab (R, QQ over ZZ))).
   $
   // Exercise: $e_M$ is one-to-one (mono). (like what we did before.) [TODO]
   We would like to show that $e_M$ is an injective function.
-  We only need to show that for any $0 != m in M$, there exists $phi : M -> hom_Ab (R, QQ over ZZ)$ such that $phi(m) != 0$. Notice that we have $ phi in homr(M, hom_Ab (R, QQ over ZZ)) iso hom_Ab (M, QQ over ZZ) $
-  as before.
-  Hence we only need to find some $phi : M -> QQ over ZZ$ in $Ab$ so that $phi(m) != 0$, which is given by @map-to-q-over-z.
+  We only need to show that for any $0 != m in M$, there exists $phi : M -> hom_Ab (R, QQ over ZZ)$ such that $phi(m) != 0$. Then since $ phi in homr(M, hom_Ab (R, QQ over ZZ)) iso hom_Ab (M, QQ over ZZ) $
+  by @tensor-hom,
+  we only need to find some $phi : M -> QQ over ZZ$ in $Ab$ so that $phi(m) != 0$, which is given by @map-to-q-over-z.
 ]