Resolutions

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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. An object \(P\) of \(\mathcal{A}\) is called a projective object if whenever the following diagram is given

   there exists at least one morphism \(P \rightarrow B\) making the following diagram

   commute.
   If for every object \(A\) of \(\mathcal{A}\) 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. An object \(I\) of \(\mathcal{A}\) is called an injective object if whenever the following diagram is given

   there exists at least one morphism \(B \rightarrow I\) making the following diagram

   commute.
   If for every object \(A\) of \(\mathcal{A}\) 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 For an object \(M\) of an abelian category \(\mathcal{A}\), we define the following.

1. A left resolution of \(M\) is a chain complex \(P_\bullet\) and an augmentation map \(\epsilon: P_0 \rightarrow M\) such that the chain complex

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

   is exact. If all the \(P_i\) are projective objects, we call this a projective resolution.

2. A right resolution of \(M\) is a cochain complex \(I^\bullet\) and an augmentation map \(\eta: M \rightarrow I^0\) such that the cochain complex

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

   is exact. 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 every object \(M\) of \(\mathcal{A}\) has a projective resolution. Similarly, if an abelian category \(\mathcal{A}\) has enough injectives, then every object \(M\) of \(\mathcal{A}\) 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 \(\epsilon_1:P_1 \rightarrow M_0\). 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 \(M_n=\ker \epsilon_n\) and obtain the following commutative diagram

Then looking at the complex obtained in the middle,

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

we obtain the identity

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

Here, the identity \(\im(\iota_{n-1}\circ\epsilon_n)=\im(\iota_{n-1})\) uses the fact that \(\epsilon_n\) is surjective, and the identity \(\ker(\epsilon_{n-1})=\ker(d_{n-1})\) uses the fact that \(\iota_{n-2}\) is injective. 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 both 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

This follows immediately from [Multilinear Algebra] §Bases, ⁋Proposition 2 and [Multilinear Algebra] §Projective, Injective, and Flat Modules, ⁋Proposition 4.

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 right adjoint preserves injective objects. Then the coextension of scalars \(\Ab \rightarrow \lMod{A}\) obtained from the ring homomorphism \(\mathbb{Z}\rightarrow A\) is a right adjoint of restriction of scalars, so injective objects in \(\Ab\) become injective objects in \(\lMod{A}\). ([Algebraic Structures] §Change of Base Ring, ⁋Proposition 7) Thus it suffices to prove that \(\Ab\) has enough injectives. For any \(A\in\Ab\), this is achieved by setting

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

and defining \(e_A:A \rightarrow I(A)\) by \(a\mapsto (f(a))_{f\in\Hom(A, \mathbb{Q}/\mathbb{Z})}\).

Uniqueness of Resolutions

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

Theorem 6 Given a projective resolution \(P_\bullet \rightarrow M\) and any \(u:M \rightarrow N\). 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, given an injective resolution \(N \rightarrow I^\bullet\) and any \(u: M \rightarrow N\), for any right resolution \(M \rightarrow J^\bullet\) there exists a chain map \(f:J^\bullet \rightarrow I^\bullet\) making the following diagram

commute.

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 chain map) We construct the \(f_n\) inductively. Since \(\varepsilon'\) is surjective and \(P_0\) is projective, for \(u\circ\varepsilon:P_0 \rightarrow N\) there exists \(f_0:P_0 \rightarrow Q_0\) such that \(\varepsilon'\circ f_0=u\circ \varepsilon\). 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, there exists \(f_n:P_n \rightarrow Q_n\) such that \(d_n^Q\circ f_n=\varphi\).

(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 \(\varepsilon'\circ g_0=u\circ\varepsilon-u\circ\varepsilon=0\). We construct a homotopy \(s_n:P_n \rightarrow Q_{n+1}\) inductively such that \(g_n=d_{n+1}^Q\circ s_n+s_{n-1}\circ d_n^P\) holds for all \(n\). 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 projectivity of \(P_0\) there exists \(s_0:P_0 \rightarrow Q_1\) such that \(d_1^Q\circ s_0=g_0\). 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 projectivity of \(P_n\) there exists \(s_n:P_n \rightarrow Q_{n+1}\) such that \(d_{n+1}^Q\circ s_n=\psi\). 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 \(\eta'\circ u:M \rightarrow I^0\) there exists \(f^0:J^0 \rightarrow I^0\) such that \(f^0\circ\eta=\eta'\circ u\). 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 this map is zero on \(\im d_J^{n-2}=\ker d_J^{n-1}\), and therefore induces a morphism to \(I^n\) via the injective morphism \(J^{n-1}/\ker d_J^{n-1}\hookrightarrow J^n\). 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 be used importantly in the next post.

Lemma 7 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 \(A'\), \(A''\) respectively. 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 of these complexes

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

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 \(i_A\) and \(\epsilon'\), so taking their direct sum we obtain \(\epsilon:P_0 \rightarrow A\). Then from §Diagram chasing, ⁋Lemma 5 we obtain the following diagram

and in particular the following diagram

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

---

References

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