Resolutions

Projective and Injective Resolutions

We defined projective and injective modules in [Multilinear Algebra] §Projective, Injective, and Flat Modules, ⁋Definition 3. Rephrasing this in the language of diagrams, we obtain the notions of projective object and injective object in a general abelian category.

Definition 1 Fix an abelian category \(\mathcal{A}\).

1. In \(\mathcal{A}\), an object \(P\) is called a projective object if whenever the following diagram

   is given, there exists at least one \(P \rightarrow B\) making the following diagram

   commute.
   If in \(\mathcal{A}\), for every object \(A\), there exists a suitable projective object \(P\) such that \(P \rightarrow A \rightarrow 0\) is exact, we say that \(\mathcal{A}\) has enough projectives.

2. In \(\mathcal{A}\), an object \(I\) is called an injective object if whenever the following diagram

   is given, there exists at least one \(B \rightarrow I\) making the following diagram

   commute.
   If in \(\mathcal{A}\), for every object \(A\), there exists a suitable injective object \(I\) such that \(0 \rightarrow A \rightarrow I\) is exact, we say that \(\mathcal{A}\) has enough injectives.

We also define the following.

Definition 2 In an abelian category \(\mathcal{A}\), for an object \(M\), we define the following.

1. A left resolution of \(M\) is, such that the following chain complex

   \[\cdots \longrightarrow P_2 \longrightarrow P_1 \longrightarrow P_0 \overset{\epsilon}{\longrightarrow} M \longrightarrow 0\]

   is exact, a chain complex \(P_\bullet\) and an augmentation map \(\epsilon: P_0 \rightarrow M\). If all the \(P_i\) are projective objects, we call this a projective resolution.

2. A right resolution of \(M\) is, such that the following cochain complex

   \[0 \longrightarrow M \overset{\eta}{\longrightarrow} I^0 \longrightarrow I^1 \longrightarrow I^2 \longrightarrow \cdots\]

   is exact, a cochain complex \(I^\bullet\) and an augmentation map \(\eta: M \rightarrow I^0\). If all the \(I^i\) are injective objects, we call this an injective resolution.

A projective object in \(\mathcal{A}\) is an injective object in \(\mathcal{A}^\op\). Similarly, if \(\mathcal{A}\) has enough projectives, then \(\mathcal{A}^\op\) has enough injectives. Also, a projective resolution of \(M\) in \(\mathcal{A}\) is the same as an injective resolution of \(M\) in \(\mathcal{A}^\op\). Therefore, it suffices to prove the following proposition only for projective resolutions.

Proposition 3 If an abelian category \(\mathcal{A}\) has enough projectives, then in \(\mathcal{A}\), every object \(M\) has a projective resolution. Similarly, if an abelian category \(\mathcal{A}\) has enough injectives, then in \(\mathcal{A}\), every object \(M\) has an injective resolution.

Proof

First, since \(\mathcal{A}\) has enough projectives, we can choose a suitable surjection \(\epsilon_0:P_0 \rightarrow M\). Let \(M_0=\ker \epsilon_0\). Then since \(\mathcal{A}\) has enough projectives, we can choose a suitable surjection \(\epsilon_1:P_1 \rightarrow M_0\). Now, drawing the composition of \(\epsilon_1: P_1 \rightarrow M_0\) and the inclusion \(\iota_0: M_0 \rightarrow P_0\) as \(d_1=\iota_0\circ\epsilon_1\) in a diagram, we obtain the following.

Continuing in this way, whenever \(\epsilon_n:P_n \rightarrow M_{n-1}\) is given, we set \(M_n=\ker \epsilon_n\) to obtain the following commutative diagram

Then looking at

\[\cdots \overset{d_3}{\longrightarrow} P_2 \overset{d_2}{\longrightarrow} P_1 \overset{d_1}{\longrightarrow} P_0 \overset{\epsilon_0}{\longrightarrow} M \longrightarrow 0\]

obtained in the middle, for \(n\geq 2\) we obtain the following equation

\[\im(d_n)=\im(\iota_{n-1}\circ\epsilon_n)=\im(\iota_{n-1})=\ker(\epsilon_{n-1})=\ker(\iota_{n-2}\circ\epsilon_{n-1})=\ker(d_{n-1})\]

and for \(n=1\) we also have \(\im(d_1)=\im(\iota_0\circ\epsilon_1)=\im(\iota_0)=\ker(\epsilon_0)\). Here, the equation \(\im(\iota_{n-1}\circ\epsilon_n)=\im(\iota_{n-1})\) uses the fact that \(\epsilon_n\) is surjective, and the equation \(\ker(\epsilon_{n-1})=\ker(d_{n-1})\) uses the fact that \(\iota_{n-2}\) is injective, respectively. Therefore \(P_\bullet\) is a projective resolution of \(M\).

One of our goals in this post is to prove that every \(A\)-module always has a projective resolution and an injective resolution. Using Proposition 3, it suffices to prove that \(\lMod{A}\) has enough projectives and enough injectives. That \(\lMod{A}\) has enough projectives is trivial.

Proposition 4 The category \(\lMod{A}\) has enough projectives.

Proof

For any \(M\in\lMod{A}\), from [Multilinear Algebra] §Bases, ⁋Proposition 2 we obtain a free module \(F\) and a surjection \(F \rightarrow M\), and from [Multilinear Algebra] §Projective, Injective, and Flat Modules, ⁋Proposition 4, \(F\) is projective.

However, since we know nothing about \(\lMod{A}^\op\), it does not follow from the above result that \(\lMod{A}\) has enough injectives. Therefore, the following proposition requires a separate proof.

Proposition 5 The category \(\lMod{A}\) has enough injectives.

Proof

One can easily show that a functor with an exact left adjoint preserves injective objects. Then, from the ring homomorphism \(\mathbb{Z}\rightarrow A\), the resulting coextension of scalars \(\Ab \rightarrow \lMod{A}\) is a right adjoint of restriction of scalars, which is an exact functor, so an injective object of \(\Ab\) becomes an injective object when sent to \(\lMod{A}\). ([Algebraic Structures] §Change of Scalars, ⁋Proposition 7) Also, if we view any \(A\)-module \(M\) as an abelian group and choose an injection \(j:M\rightarrow I\) into an injective object \(I\) of \(\Ab\), the \(A\)-module homomorphism \(M\rightarrow\Hom_\mathbb{Z}(A,I)\) obtained by adjunction is given by \(m\mapsto(a\mapsto j(am))\), and evaluating this at \(a=1\) yields \(j\), so this is also an injection. Therefore, for the desired proof, it suffices to prove that \(\Ab\) has enough injectives. This is done by defining, for any \(G\in\Ab\),

\[I(G)=\prod_{f\in\Hom_\Ab(G, \mathbb{Q}/\mathbb{Z})} \mathbb{Q}/\mathbb{Z}\]

and \(e_G:G \rightarrow I(G)\) by \(g\mapsto (f(g))_{f\in\Hom(G, \mathbb{Q}/\mathbb{Z})}\).

Uniqueness of Resolutions

Meanwhile, the uniqueness of projective resolutions and injective resolutions is obtained from the following stronger theorem.

Theorem 6 Let a projective resolution \(P_\bullet \rightarrow M\) and any \(u:M \rightarrow N\) be given. Then for any left resolution \(Q_\bullet \rightarrow N\), there exists a chain map \(f:P_\bullet \rightarrow Q_\bullet\) making the following diagram

commute, uniquely up to homotopy.
Similarly, let an injective resolution \(N \rightarrow I^\bullet\) and any \(u: M \rightarrow N\) be given. Then for any right resolution \(M \rightarrow J^\bullet\), there exists a chain map \(f:J^\bullet \rightarrow I^\bullet\) making the following diagram

commute, uniquely up to homotopy.

Proof

First we prove the first claim. Write the augmentations of the two resolutions as \(\varepsilon:P_0 \rightarrow M\) and \(\varepsilon':Q_0 \rightarrow N\), respectively.

(Existence of a chain map) We construct the \(f_n\) inductively. Since \(\varepsilon'\) is surjective and \(P_0\) is projective, for \(u\circ\varepsilon:P_0 \rightarrow N\) we have \(\varepsilon'\circ f_0=u\circ \varepsilon\) for some \(f_0:P_0 \rightarrow Q_0\). Now suppose \(f_0,\ldots,f_{n-1}\) have been constructed to make the given diagram commute, and consider the composition \(\varphi=f_{n-1}\circ d_n^P:P_n \rightarrow Q_{n-1}\). Then for \(n\geq 2\),

\[d_{n-1}^Q\circ\varphi=d_{n-1}^Q\circ f_{n-1}\circ d_n^P=f_{n-2}\circ d_{n-1}^P\circ d_n^P=0\]

and for \(n=1\) we also have \(\varepsilon'\circ f_0\circ d_1^P=u\circ\varepsilon\circ d_1^P=0\). Therefore \(\im\varphi\) is contained in \(\ker d_{n-1}^Q\) (or \(\ker\varepsilon'\) when \(n=1\)), which equals \(\im d_n^Q\) since \(Q_\bullet \rightarrow N\) is a resolution. Then since \(P_n\) is projective and \(d_n^Q:Q_n \rightarrow \im d_n^Q\) is surjective, we have \(d_n^Q\circ f_n=\varphi\) for some \(f_n:P_n \rightarrow Q_n\).

(Uniqueness up to homotopy) Suppose \(f,f'\) are both chain maps making the given diagram commute, and let \(g=f-f'\). Then \(g\) is a chain map satisfying \(\varepsilon'\circ g_0=u\circ\varepsilon-u\circ\varepsilon=0\). For \(g_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P\) to hold for all \(n\), we construct a homotopy \(s_n:P_n \rightarrow Q_{n+1}\) inductively. Here \(s_{-1}=0\).

First, from \(\varepsilon'\circ g_0=0\) we have \(\im g_0\subseteq \ker\varepsilon'=\im d_1^Q\), so by the projectivity of \(P_0\), we have \(d_1^Q\circ s_0=g_0\) for some \(s_0:P_0 \rightarrow Q_1\). Now suppose \(s_0,\ldots,s_{n-1}\) have been constructed and let \(\psi=g_n-s_{n-1}\circ d_n^P\). Then

\[d_n^Q\circ\psi=d_n^Q\circ g_n-(d_n^Q\circ s_{n-1})\circ d_n^P=g_{n-1}\circ d_n^P-(g_{n-1}-s_{n-2}\circ d_{n-1}^P)\circ d_n^P=s_{n-2}\circ d_{n-1}^P\circ d_n^P=0\]

so \(\im\psi\subseteq\ker d_n^Q=\im d_{n+1}^Q\), and again by the projectivity of \(P_n\), we have \(d_{n+1}^Q\circ s_n=\psi\) for some \(s_n:P_n \rightarrow Q_{n+1}\). Then by definition \(g_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P\).

Now we prove the second claim. This is obtained by dualizing the proof of the first claim verbatim. Let the augmentations of the two resolutions be \(\eta:M \rightarrow J^0\) and \(\eta':N \rightarrow I^0\). Since \(\eta\) is injective and \(I^0\) is an injective object, for \(\eta'\circ u:M \rightarrow I^0\) we have \(f^0\circ\eta=\eta'\circ u\) for some \(f^0:J^0 \rightarrow I^0\). Suppose \(f^0,\ldots,f^{n-1}\) have been constructed inductively, and consider the composition \(d_I^{n-1}\circ f^{n-1}:J^{n-1} \rightarrow I^n\). By the same computation as above, on \(\im d_J^{n-2}=\ker d_J^{n-1}\) this map is \(0\), and therefore, via the injective morphism \(J^{n-1}/\ker d_J^{n-1}\hookrightarrow J^n\), it induces a morphism to \(I^n\). Since \(I^n\) is an injective object, we extend this to all of \(J^n\) to obtain \(f^n:J^n \rightarrow I^n\), and by this construction \(f^n\circ d_J^{n-1}=d_I^{n-1}\circ f^{n-1}\) holds.

Finally, we conclude by proving the following lemma, which will play an important role in the next post.

Lemma 7 Suppose we are given the following short exact sequence

\[0 \longrightarrow A'\overset{i}{\longrightarrow}A\overset{p}{\longrightarrow}A'' \longrightarrow 0\]

and projective resolutions \(P_\bullet'\), \(P_\bullet''\) of \(A'\), \(A''\). Then the chain complex \(P_\bullet\) defined by \(P_n=P_n'\oplus P_n''\) is a projective resolution of \(A\), and there exists an exact sequence

\[0 \rightarrow P' \rightarrow P \rightarrow P'' \rightarrow 0\]

between these complexes.

Proof

First, drawing the given situation in a diagram, we obtain the following.

Now from the condition that \(P_0''\) is projective, we can define \(P_0'' \rightarrow A\). On the other hand, \(P_0' \rightarrow A\) is already given as the composition of \(i\) and \(\epsilon'\), so taking their direct sum we obtain \(\epsilon:P_0 \rightarrow A\). Now from §Diagram chasing, ⁋Theorem 6 (The snake lemma) we obtain the following diagram

and in particular the following diagram

Repeating this process, we obtain \(P_\bullet\).

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References

[Wei] C.A. Weibel. An Introduction to Homological Algebra. Cambridge Studies in Advanced Mathematics. Cambridge University Press, 1995.