# Friday, February 05 > See missing first 10m > Recall the definition of injectives. :::{.remark} Over a PID, divisible is equivalent (?) to injective as a module. ::: :::{.example title="?"} $\QQ$ is divisible, and thus an injective $\ZZ\dash$module. Similarly $\QQ/\ZZ \mapstofrom [0, 1) \intersect \QQ$. ::: :::{.example title="?"} Let $p\in \ZZ$ be prime, then $\ZZ[{1\over p}] \subseteq \QQ$ has elements of the form \( \sum {a_i \over p^{n_i} } \), and is not divisible. On the other hand, $\ZZ_{p^ \infty }\da \ZZ[{1\over p}]/\ZZ \mapstofrom \ZZ[{1\over p}] \intersect [0, 1)$ is divisible since $p^n \qty{ a\over p^n } = a \in \ZZ$, which equals zero in $\ZZ_{p^{\infty }}$. To solve $xr = a/p^n$ with $r,a \in \ZZ$ and $r\neq 0$, first assume $\gcd(r, p) = 1$ by just dividing through by any common powers of $p$. This amounts to solving $1 = sr tp^n$ where $s, t\in \ZZ$: \[ {a\over p^n} &= sr \qty{a \over p^n} + tp^n\qty{a \over p^n} \\ &= \qty{ sa \over p^n} r \\ &\da xr \in \ZZ_{p^{\infty }} .\] ::: :::{.fact} Every injective abelian group is isomorphic to a direct sum of copies of $\QQ$ and $\ZZ_{p^{\infty }}$ for various primes $p$. ::: :::{.example title="?"} $\QQ/\ZZ \cong \bigoplus _{p\text{ prime}} \ZZ_{p^{\infty }}$. To prove this, do induction on the number of prime factors in the denominator. ::: :::{.exercise title="2.3.2"} $\Ab = \mods{\ZZ}$ has enough injectives. ::: :::{.remark} As a consequence, $\modr$ has enough injectives for *any* ring $R$. ::: ## Transferring Injectives Between Categories Next we'll use our background in projectives to deduce analogous facts for injectives. :::{.definition title="Opposite Category"} Let \( \mathcal{A} \) be any category, then there is an opposite/dual category \( \mathcal{A}\op \) defined in the following way: - $\Ob(\mathcal{A}\op) = \Ob(\mathcal{A})$ - $A\to B\in \Mor(\mathcal{A}) \implies B\to A \in \Mor(\mathcal{A}\op)$, so \[ \Hom_{\mathcal{A}}(A, B) &\mapstofrom \Hom_{ \mathcal{A}\op}(B, A) \\ f &\mapstofrom f\op .\] - We require that if $A \mapsvia{f} B \mapsvia{g} C$ in \( \mathcal{A} \), then $f\op \circ g\op = (g\circ f)\op$ where $C \mapsvia{g\op} B \mapsvia{f\op} A$. - $\one_{A}\op = \one_{A}$ in \( \mathcal{A}\op \). ::: :::{.warnings} Thinking of these as functions won't quite work! For $f:A\to B$, there may not be any map $B\to A$ -- you'd need it to be onto to even define such a thing, and if it's not injective there are many choices. Note that initials and terminals are swapped, and since $0$ is both. Counterintuitively, $A \to 0 \to B$ is $0$, which maps to $B \to 0 \to A = 0\op$. ::: :::{.remark} Note that $(\wait)\op$ switches - Monics and epis, - Initial and terminal objects, - Kernels and cokernels. Moreover, \( \mathcal{A} \) is abelian if and only if \( \mathcal{A}\op \) is abelian. ::: :::{.definition title="Contravariant Functors"} A **contravariant functor** $F: \mathcal{C}\to \mathcal{D}$ is a *covariant* functor \( \mathcal{C}\op \to \mathcal{D} \). \begin{tikzcd} {C_1} && {C_2} && {C_2} && {C_1} && {FC_2} && {FC_1} \\ && {\mathcal{C}} && {\mathcal{C}\op} && {\mathcal{C}\op} && {\mathcal{D}} \arrow["f", from=1-1, to=1-3] \arrow["{f\op}", from=1-5, to=1-7] \arrow["{F(f)}", from=1-9, to=1-11] \arrow[squiggly, from=2-3, to=2-5] \arrow[squiggly, from=2-7, to=2-9] \end{tikzcd} In particular, $F(\one) = \one$ and $F(gf) = F(f) F(g)$ > [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzIsMCwiQ18yIl0sWzAsMCwiQ18xIl0sWzQsMCwiQ18yIl0sWzYsMCwiQ18xIl0sWzgsMCwiRkNfMiJdLFsxMCwwLCJGQ18xIl0sWzIsMSwiXFxtYXRoY2Fse0N9Il0sWzQsMSwiXFxtYXRoY2Fse0N9Xlxcb3AiXSxbNiwxLCJcXG1hdGhjYWx7Q31eXFxvcCJdLFs4LDEsIlxcbWF0aGNhbHtEfSJdLFsxLDAsImYiXSxbMiwzLCJmXlxcb3AiXSxbNCw1LCJGKGYpIl0sWzYsNywiIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoic3F1aWdnbHkifX19XSxbOCw5LCIiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJzcXVpZ2dseSJ9fX1dXQ==) ::: :::{.example title="?"} $\Hom_R(\wait, A): \modr \to \Ab$ is a contravariant functor in the first slot. ::: :::{.definition title="Left-Exact Functors"} A contravariant functor $F: \mathcal{A} \to \mathcal{B}$ between abelian categories is **left exact** if and only if the corresponding covariant functor $F: \mathcal{A}\op \to \mathcal{B}$: That is, SESs in \( \mathcal{A} \) get mapped to long left-exact sequences in \( \mathcal{B} \) : % https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIkEiXSxbNCwwLCJCIl0sWzYsMCwiQyJdLFs4LDAsIjAiXSxbMCwyLCIwIl0sWzIsMiwiRkMiXSxbNCwyLCJGQiJdLFs2LDIsIkZBIl0sWzgsMiwiXFxjZG90cyJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs1LDZdLFs2LDddLFs3LDhdLFs4LDldLFsyLDcsIkYoXFx3YWl0KSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6InNxdWlnZ2x5In19fV1d \begin{tikzcd} 0 && A && B && C && 0 \\ \\ 0 && FC && FB && FA && \cdots \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow["{F(\wait)}"{description}, squiggly, from=1-5, to=3-5] \end{tikzcd} ::: :::{.lemma title="?"} If \( \mathcal{A} \) is abelian and $A \in \mathcal{A}$, then the following are equivalent: - $A$ is injective in \( \mathcal{A} \). - $A$ is projective in \( \mathcal{A}\op \). - The contravariant functor $\Hom_{\mathcal{A}}(\wait, A)$ is exact. ::: :::{.lemma title="?"} If an abelian category \( \mathcal{A} \) has enough injectives, then every $M\in \mathcal{A}$ has an injective resolution: \[ 0 \to M \to I^0 \to I^1 \to \cdots .\] which is an exact cochain complex with each $I^n$ injective. There is a version of the comparison lemma that is proved in roughly the same way as for projective resolutions. ::: Next up: how to transport injective resolutions in $\mods{\ZZ}$ to $\rmod$. :::{.observation} If \( A\in \Ab \) and \( N \in \rmod \) then \( \Hom_{\Ab}(N, A) \in \modr \) in the following way: taking $f: N\to A$ and $r\in R$, define a right action $(f\cdot r)(n) \da f(rn)$. ::: :::{.exercise title="?"} Check that this is a morphism of abelian groups, that this yields a module structure, along with other details. For noncommutative rings, it's crucial that the $r$ is on the right in the action and on the left in the definition. ::: :::{.lemma title="?"} If $M \in \modr$, then the following natural map \( \tau \) is an isomorphism of abelian groups for each $A\in \Ab$: \[ \tau: \Hom_{\Ab}(\forget(M), A) &\to \Hom_{\modr}(M, \Hom_{\Ab}(R, A)) \\ f &\mapsto \tau(f)(m)(r) \da f(mr) ,\] where $m\in M$ and $r\in R$ and \( \forget: \modsright{R}\to \modsright{\ZZ} \) is a forgetful functor. Note that $R$ is a left \(R\dash\)module, so the hom in the RHS is a right \(R\dash\)module and the hom makes sense. :::{.exercise title="?"} Check the details here, particularly that the module structures all make sense. ::: There is a map \( \mu \) going the other way: $\mu(g)(m) = g(m)(1_R)$ for $m\in M$. ::: :::{.remark} A quick look at why these maps are inverses: \[ \mu( \tau(f)) &= (\tau f)(m)(1_R) \\ &= f(m\cdot 1) \\ &= f(m) .\] Conversely, \[ \tau(\mu(g))(m)(r) &= (\mu g)( mr) \\ &= g(mr)(1) \\ &= g(m\cdot r) && \text{since } g \in \Mor_{\rmod} \\ &= g(m)(r\cdot 1) && \text{by observation earlier} \\ &= g(m)(r) .\] ::: :::{.remark} The $?$ functor in the lemma will be the forgetful functor applied to $M$, yielding an adjoint pair. :::