# Homological Algebra Problem Sets

Prove the 5-lemma. Suppose the following rows are exact:

• Show that if $$f_2, f_4$$ are monic and $$f_1$$ is an epi, then $$f_3$$ is monic.
• Show that if $$f_2, f_4$$ are epi and $$f_5$$ is monic, then $$f_3$$ is an epi.
• Conclude that if $$f_1, f_2, f_4, f_5$$ are isomorphisms then $$f_4$$ is an isomorphism.

Appealing to the Freyd-Mitchell Embedding Theorem, we proceed by chasing elements. For this part of the result, only the following portion of the diagram is relevant, where monics have been labeled with “$$\hookrightarrow$$” and the epis with “$$\twoheadrightarrow$$”:

It suffices to show that $$f_3$$ is an injection, and since these can be thought of as $$R{\hbox{-}}$$module morphisms, it further suffices to show that $$\ker f_3 = 0$$, so $$f_3(x) = 0 \implies x = 0$$. The following outlines the steps of the diagram chase, with references to specific square and element names in a diagram that follows:

• Suppose $$x\in A_3$$ and $$f(x) = 0\in B_3$$.
• Then under $$A_3 \to B_3 \to B_4$$, $$x$$ maps to zero.
• Letting $$y_1$$ be the image of $$x$$ under $$A_3 \to A_4$$, commutativity of square 1 and injectivity of $$f_4$$ forces $$y_1 = 0$$.
• Exactness of the top row allows pulling this back to some $$y_2 \in A_2$$.
• Under $$A_2 \to B_2$$, $$y_2$$ maps to some unique $$y_3\in B_2$$, using injectivity of $$f_2$$.
• Commutativity of square 2 forces $$y_3 \to 0$$ under $$B_2 \to B_3$$.
• Exactness of the bottom row allows pulling this back to some $$y_3\in B_1$$
• Surjectivity of $$f_1$$ allows pulling this back to some $$y_5 \in A_1$$.
• Commutativity of square 3 yields $$y_5\mapsto y_2$$ under $$A_1\to A_2$$ and $$y_5 \mapsto x$$ under $$A_1 \to A_2 \to A_3$$.
• But exactness in the top row forces $$y_5 \mapsto 0$$ under $$A_1 \to A_2 \to A_3$$, so $$x=0$$.

Similarly, it suffices to consider the following portion of the diagram:

We’ll proceed by starting with an element in $$B_3$$ and constructing an element in $$A_3$$ that maps to it. We’ll complete this in two steps: first by tracing around the RHS rectangle with corners $$B_3, B_5, A_5, A_3$$ to produce an “approximation” of a preimage, and second by tracing around the LHS square to produce a “correction term”. Various names and relationships between elements are summarized in a diagram following this argument.

Step 1 (the right-hand side approximation):

• Let $${\color{red}y}\in B_3$$ and $$y_1$$ be its image under $$B_3\to B_4$$.
• By exactness of the bottom row, under $$B_4 \to B_5$$, $$y_1\mapsto 0$$.
• By surjectivity of $$f_4$$, pull $$y_1$$ back to an element $$y_2 \in A_4$$.
• By commutativity of square 1, $$y_2\mapsto 0$$ under $$A_4\to A_5 \to B_5$$.
• By injectivity of $$f_5$$, the preimage of zero must be zero and thus $$y_2\mapsto 0$$ under $$A_4\to A_5$$.
• Using exactness of the top row, pull $$y_2$$ back to obtain some $${\color{blue}y_3} \in A_3$$

Step 2 (the left-hand correction term):

• Let $$z$$ be the image of $$\color{blue}y_3$$ under $$A_3\to B_3$$, noting that $$z\neq y$$ in general.
• By commutativity of square 2, $$z\mapsto y_1$$ under $$B_3\to B_4$$
• Thus $$z-y\mapsto y_1 - y_1 = 0$$ under $$B_3\to B_4$$, using that $$d(z-y) = d(z) - d(y)$$ since these are $$R{\hbox{-}}$$module morphisms.
• By exactness of the bottom row, pull $$z-y$$ back to some $$\zeta_1\in B_2$$.
• By surjectivity of $$f_2$$, pull this back to $$\zeta_2 \in A_2$$. Note that by construction, $$\zeta_2 \mapsto z-y$$ under $$A_2\to B_2\to B_3$$.
• Let $$\zeta_3$$ be the image of $$\zeta_2$$ under $$A_2\to A_3$$.
• By commutativity of square 3, $$\zeta_4 \mapsto z-y$$ under $$A_3\to B_3$$.
• But then $${\color{blue}y_3} - \zeta_3 \mapsto z - (z-y) = {\color{red}y}$$ under $$A_3\to B_3$$ as desired.

Given the previous two result, if the outer maps are isomorphisms then $$f_3$$ is both monic and epi. Using a technical fact that monic epis are isomorphisms in a category $$\mathcal{C}$$ if and only if $$\mathcal{C}$$ is balanced and that all abelian categories are balanced, $$f_3$$ is isomorphism.

Let $$C$$ be a chain complex. Show that $$C$$ is split if and only if there are $$R{\hbox{-}}$$module decompositions \begin{align*} C_n & \cong Z_n \oplus B_n' \\ Z_n &= B_n \oplus H_n' .\end{align*} Show that $$C$$ is split exact if and only if $$H_n' = 0$$.

For this problem, we’ll use the fact that if $$d$$ is an epimorphism, it satisfies the right-cancellation property: if $$f\circ d = g\circ d$$, then $$f=g$$. In particular, if $$d_n = d_ns_{n-1}d_n$$ with $$d_n: C_n\to C_{n-1}$$ surjective and $$s_{n-1}: C_{n-1}\to C_n$$, we can conclude $$\one_{C_n} = d_ns_{n-1}$$. We’ll also use the fact that if we have a SES in any abelian category $$\mathcal{A}$$, then the following are equivalent:

• The sequence is split on the left.
• The sequence is split on the right.
• The middle term is isomorphic to the direct sum of the outer terms.

Fixing notation, we’ll write $$C \coloneqq(\cdots \to C_{n+1} \xrightarrow{{{\partial}}_{n+1}} C_n \xrightarrow{{{\partial}}_n} \cdots)$$, and we’ll use concatenation $$fg$$ to denote function composition $$f \circ g$$.

$$\implies$$:

Suppose $$C$$ is split, so we have maps $$\left\{{s_n}\right\}$$ such that $${{\partial}}_n = {{\partial}}_n s_{n-1} {{\partial}}_n$$.

The short exact sequence \begin{align*} 0 \to Z_n \xrightarrow{\iota} C_n \xrightarrow{{{\partial}}_n} B_{n-1} \to 0 \end{align*} admits a right-splitting $$f: B_{n-1}\to C_n$$, and thus there is an isomorphism \begin{align*} C_n \cong Z_n \oplus B_n' \coloneqq Z_n \oplus B_{n-1} .\end{align*}

That this sequence is exact follows from the fact that it can be written as \begin{align*} 0 \to \ker {{\partial}}_n \hookrightarrow C_n \overset{{{\partial}}_n}\twoheadrightarrow\operatorname{im}{{\partial}}_n \to 0 .\end{align*}

We proceed by constructing the splitting $$f$$. Noting that $$s_{n-1}: C_{n-1} \to C_n$$ and $$B_{n-1} \leq C_{n-1}$$, the claim is that its restriction $$f\coloneqq{ \left.{{s_{n-1}}} \right|_{{B_{n-1}}} }$$ works. It suffices to show that $$(C_n \xrightarrow{{{\partial}}_n} B_{n-1} \xrightarrow{f} C_n)$$ composes to the identity map $$\one_{C_n}$$. This follows from the splitting assumption, along with right-cancellability since $${{\partial}}_n$$ is surjective onto its image:

\begin{align*} {{\partial}}_n = {{\partial}}_n s_{n-1} {{\partial}}_n \overset{\text{right-cancel ${{\partial}}_n$}}{\implies} \one_{C_n} = {{\partial}}_n s_{n-1} \coloneqq{{\partial}}_n f .\end{align*}

The SES \begin{align*} 0 \to B_n \overset{\iota_{BZ}}\hookrightarrow Z_n \overset{\pi}{\twoheadrightarrow} \frac{Z_n}{B_n} \to 0 \end{align*} admits a left-splitting $$f: Z_n\to B_n$$, and thus there is an isomorphism \begin{align*} Z_n \cong B_n \oplus H_n' \coloneqq B_n \oplus H_n(C) \coloneqq B_n \oplus {Z_n \over B_n} .\end{align*}

We proceed by again constructing the splitting $$f:Z_n\to B_n$$. The situation is summarized in the following diagram:

So a natural candidate for the map $$f$$ is the composition \begin{align*} Z_n \xrightarrow{\iota_Z} C_n \xrightarrow{s_{n}} C_{n+1} \xrightarrow{{{\partial}}_{n+1}} B_n ,\end{align*} so $$f \coloneqq{{\partial}}_{n+1} s_{n} \iota_Z$$. We can simplify this slightly by regarding $$Z_n \leq C_n$$ as a submodule to suppress $$\iota_Z$$, and identify $$s_{n}$$ with its restriction to $$Z_n$$ to write $$f \coloneqq{{\partial}}_{n+1} s_{n}$$. The claim is then that $$f\iota_{BZ} = \one_{B_n}$$. Anticipating using the fact that $${{\partial}}_{n+1} = {{\partial}}_{n+1} s_n {{\partial}}_{n+1}$$, we post-compose with $${{\partial}}_{n+1}$$ and compute: \begin{align*} f \iota_{BZ} {{\partial}}_{n+1} &= (C_{n+1} \xrightarrow{{{\partial}}_{n+1}} B_n \xrightarrow{\iota_{BZ}} Z_n \xrightarrow{\iota_Z} C_n \xrightarrow{s_n} C_{n+1} \xrightarrow{{{\partial}}_{n+1}} C_n ) \\ &= (C_{n+1} \xrightarrow{{{\partial}}_{n+1}} B_n \xrightarrow{{ \left.{{s_n}} \right|_{{B_n}} } } C_{n+1} \xrightarrow{{{\partial}}_{n+1}} ) \\ &= {{\partial}}_{n+1} s_n {{\partial}}_{n+1} \\ &= {{\partial}}_{n+1} ,\end{align*} where in the last step we’ve used the splitting hypothesis and the fact that it remains true when everything is restricted to the submodule $$B_n \leq C_n$$. Using surjectivity of $${{\partial}}_{n+1}$$ onto $$B_n$$, we can now conclude as before: \begin{align*} f \iota_{BZ} {{\partial}}_{n+1} = {{\partial}}_{n+1} \overset{\text{right-cancel ${{\partial}}_n$}}{\implies} f\iota_{BZ} = \one_{B_n} .\end{align*}

Show that $$C$$ is a split exact chain complex if and only if $$\one_C$$ is nullhomotopic.

$$\impliedby$$:

$$C$$ is split: Suppose $$\one_{C}$$ is nullhomotopic, so that there exist maps \begin{align*} s_n: C_n \to C_{n+1} && \one_{C_n} = {{\partial}}_{n+1} s_n + s_{n-1} {{\partial}}_n .\end{align*} We then have the following situation:

Here the nullhomotopy is outlined in red, and the map relevant to the splitting in green. Note that $$s_n: C_n \to C_{n+1}$$ is a candidate for a splitting, we just need to show that $${{\partial}}_{n+1} = {{\partial}}_{n+1} s_n {{\partial}}_{n+1}$$. We can proceed by post-composing the LHS with the identity $$\one_C$$, which allows us to substitute in the nullhomotopy: \begin{align*} {{\partial}}_{n+1} &= \one_{C_n} {{\partial}}_{n+1} \\ &= \qty{ {{\partial}}_{n+1} s_n + s_{n-1} {{\partial}}_n } {{\partial}}_{n+1} \\ &= {{\partial}}_{n+1} s_n {{\partial}}_{n+1} + s_{n-1} {{\partial}}_n {{\partial}}_{n+1} \\ &= {{\partial}}_{n+1} s_n {{\partial}}_{n+1} + s_{n-1} {\color{red} 0} && \text{since } {{\partial}}^2 = 0\\ &= {{\partial}}_{n+1} s_n {{\partial}}_{n+1} .\end{align*}

$$C$$ is exact: This follows from the fact that since $$\one_C = {{\partial}}s + s{{\partial}}$$ are equal as maps of chain complexes, the images $$D_1 \coloneqq\one_C(C)$$ and $$D_2 \coloneqq({{\partial}}s + s {{\partial}})(C)$$ are equal as chain complexes and have equal homology. Evidently $$D_1 = C$$, and on the other hand, each graded piece $$(D_2)_n$$ only consists of boundaries coming from various pieces of $$C$$, since $${{\partial}}s + s{{\partial}}$$ necessarily lands in the images of the maps $${{\partial}}_n$$. Thus $$C_n(D_2) \subseteq B_n(D_2) = \emptyset$$, i.e. there are no chains (or cycles) in $$D_2$$ which are not boundaries, and thus $$H_n(D_2) \coloneqq Z_n(D_2)/ B_n(D_2) = 0$$ for all $$n$$. We can thus conclude that $$C = D_2 \implies H(C) = H(D_2) = 0$$, so $$C$$ must be exact.

$$\implies$$: Suppose $$C$$ is split. Then by exercise 1.4.2, we have $$R{\hbox{-}}$$module decompositions \begin{align*} C_n &\cong Z_n \oplus B_{n-1} \\ Z_n &\cong B_n \oplus H_n \\ \\ \implies C_n &\cong B_n \oplus B_{n-1} \oplus H_n .\end{align*}

Supposing further that $$C$$ is exact, we have $$H_n = 0$$, and thus $$C_n \cong B_n \oplus B_{n-1}$$. We first note that in this case, we can explicitly write the differential $${{\partial}}_n$$. Letting $$\one_n$$ denote the identity on $$C_n$$, where by abuse of notation we also write this for its restriction to any submodules, we have:

We can thus write $${{\partial}}_n$$ as the matrix \begin{align*} {{\partial}}_n = \begin{bmatrix} 0 & \one_n \\ 0 & 0 \end{bmatrix} .\end{align*}

Similarly using this decomposition, we can construct a map $$s_n: C_{n} \to C_{n+1}$$:

We can write this as the following matrix: \begin{align*} s_n = \begin{bmatrix} 0 & 0 \\ \one_n & 0 \end{bmatrix} .\end{align*}

We can now verify that $$s_n$$ is a nullhomotopy from a direct computation: \begin{align*} {{\partial}}_{n+1} s_n + s_{n-1} {{\partial}}_n &= \begin{bmatrix} 0 & \one_{n+1} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ \one_n & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ \one_{n-1} & 0 \end{bmatrix} \begin{bmatrix} 0 & \one_{n} \\ 0 & 0 \end{bmatrix} \\ &= \begin{bmatrix} \one_n & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & \one_{n-1} \end{bmatrix} \\ &= \begin{bmatrix} \one_{B_n} & 0 \\ 0 & \one_{B_{n-1}} \end{bmatrix} \\ &= \one_{C_n} ,\end{align*} expressed as a map $$B_n \oplus B_{n-1} \to B_n \oplus B_{n-1}$$.

Show that chain homotopy classes of maps form a quotient category $$K$$ of $$\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$ and that the functors $$H_n$$ factor through the quotient functor $$\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}}) \to K$$ using the following steps:

• Show that chain homotopy equivalence is an equivalence relation on $$\left\{{f:C\to D {~\mathrel{\Big\vert}~}f\text{ is a chain map}}\right\}$$. Define $$\mathop{\mathrm{Hom}}_K(C, D)$$ to be the equivalence classes of such maps and show that it is an abelian group.

• Let $$f \simeq g: C\to D$$ be two chain homotopic maps. If $$u: B\to C, v:D\to E$$ are chain maps, show that $$vfu, vgu$$ are chain homotopic. Deduce that $$K$$ is a category when the objects are defined as chain complexes and the morphisms are defined as in (1).

• Let $$f_0, f_1, g_0, g_1: C\to D$$ all be chain maps such that each pair $$f_i \simeq g_i$$ are chain homotopic. Show that $$f_0 + f_1$$ is chain homotopic to $$g_0 + g_1$$. Deduce that $$K$$ is an additive category and $$\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})\to K$$ is an additive functor.

• Is $$K$$ an abelian category? Explain.

Try at least two parts.

\envlist


Chain homotopy equivalence defines an equivalence relation on $$\mathop{\mathrm{Hom}}_{\mathsf{Ch}(\mathcal{A})}(A, B)$$.

We recall that for morphisms $$f,g \in \mathop{\mathrm{Hom}}_{\mathsf{Ch}(\mathcal{A} )}(A, B)$$, we have $$f \simeq g \iff f-g \simeq 0 \iff \exists s: A \to B[1]$$ such that $${{\partial}}^B s + s{{\partial}}^A = f-g$$.

• Reflexive: We want to show that $$f \simeq f$$, i.e. $$f-f = 0 \simeq 0$$. Producing the map $$s = 0$$ works, since $${{\partial}}s + s{{\partial}}= {{\partial}}0 + 0{{\partial}}= 0$$.

• Symmetric: Suppose $$f\simeq g$$, so there exists an $$s$$ such that $${{\partial}}s + s {{\partial}}= f-g$$. Then taking $$s' \coloneqq-s$$ produces a chain homotopy $$g-f\simeq 0$$, since we can write \begin{align*} {{\partial}}s' + s' {{\partial}} &= {{\partial}}(-s) + (-s) {{\partial}}\\ &= -{{\partial}}s - s{{\partial}}\\ &= -( {{\partial}}s + s {{\partial}}) \\ &= -(f-g) \\ &= g-f .\end{align*}

• Transitive: Suppose $$f\simeq g$$ and $$g\simeq h$$, we want to show $$f\simeq h$$. By assumption we have maps $$s, s'$$ such that $${{\partial}}s + s{{\partial}}= f-g$$ and $${{\partial}}s' + s'{{\partial}}= g-h$$, so set $$s'' \coloneqq s + s'$$. We can then check that this is a chain homotopy from $$f$$ to $$h$$: \begin{align*} {{\partial}}s'' + s'' {{\partial}} &= {{\partial}}(s+ s') + (s + s'){{\partial}}\\ &= ({{\partial}}s + s{{\partial}}) + ({{\partial}}s' + s' {{\partial}}) \\ &= (f-g) + (g-h) \\ &= f-h ,\end{align*} where we’ve used that abelian categories are enriched over abelian groups, so we have a commutative and associative addition on homs.

$$( \mathop{\mathrm{Hom}}_K(A, B), \oplus) \in {\mathsf{Ab}}$$, where we define an addition on equivalence classes by \begin{align*} [f] \oplus [g] \coloneqq[f + g] .\end{align*}

We check the group axioms directly:

• Closure under operation: We can check that $$[f] \oplus [g] \coloneqq[f+g] \coloneqq[g']$$ makes sense, since $$g' \coloneqq f+g \in \mathop{\mathrm{Hom}}_{\mathsf{Ch}(\mathcal{A})}(A, B)$$, making $$[g']$$ a well-defined equivalence class of maps in $$\mathop{\mathrm{Hom}}_K(A, B)$$.

• Two-sided Identities: The equivalence class $$[0] \coloneqq\left\{{f\in \mathop{\mathrm{Hom}}_{\mathsf{Ch}(\mathcal{A} )}(A, B) {~\mathrel{\Big\vert}~}f\simeq 0}\right\}$$ serves as an identity, where we take the zero map as a representative. This follows from the fact that \begin{align*} [0] \oplus [f] \coloneqq[0 + f] = [f] = [f] \oplus [0] .\end{align*}

• Associativity: This again follows from the abelian group structure on the original hom: \begin{align*} [f] \oplus \qty{ [g] \oplus [h] } = [f + (g+h)] = [(f+g) + h] = \qty{ [f] \oplus [g] } \oplus [h] .\end{align*}

• Two-sided inverses: Let $$\ominus[f] \coloneqq[-f]$$ be a candidate for the inverse with respect to $$\oplus$$. To see that this works, we have \begin{align*} [f] \oplus ( \ominus [f] ) \coloneqq[f] \oplus [-f] = [f-f] = [0] ,\end{align*} and a similar calculation goes through to show it’s also a left-sided inverse.

• Well-definedness: We need to show that if $$[f] = [f']$$ and $$[g] = [g']$$, the sums agree, so $$[f] \oplus [g] = [f'] \oplus [g']$$. We’ll use the fact that $$[f] = [f'] \iff f-f' \simeq 0$$ (and similarly for $$g$$), so we can compute: \begin{align*} \qty{ [f] \oplus [g] } \ominus \qty{ [f'] \oplus [g'] } &\coloneqq[f + g] \ominus [f' - g'] \\ &\coloneqq[(f+g) - (f'-g') ] \\ &= [(f-f') + (g-g')] \\ &= [f-f'] \oplus [g-g'] \\ &= [0] \oplus [0] && \text{using } f\simeq f',~ g\simeq g'\\ &\coloneqq[0] .\end{align*}

We have the following situation, where the double-arrows denote chain homotopies:

We want to show that $$vgu \simeq vfu$$, or equivalently that $$vgu - vfu \simeq 0$$. This is immediate: \begin{align*} vgu - vfu = v(gu - fu) = v(f - g)u \simeq v0u = 0 .\end{align*}

Alternatively, as an explicit computation, if we assume $$f\simeq g$$ then there is a nullhomotopy $$s$$ for $$f-g$$, in which case the map $$s' \coloneqq vsu$$ works as a nullhomotopy for $$vfu - vgu$$: \begin{align*} vfu - vgu &= v(f-g)u = v({{\partial}}s + s {{\partial}}) u \\ &= {\color{red} v{{\partial}}} s u + vs {\color{red} {{\partial}}u } \\ &= {\color{blue} {{\partial}}v} su + v s {\color{blue} u {{\partial}}} && \text{since $u, v$ are chain maps} \\ &\coloneqq{{\partial}}s' + s'{{\partial}} .\end{align*}

We can now define a composition map on $$\mathsf{K}$$: \begin{align*} \circ: \mathop{\mathrm{Hom}}_K(A_1, A_2) \times\mathop{\mathrm{Hom}}_K(A_2, A_3) &\to \mathop{\mathrm{Hom}}_K(A_1, A_3) \\ ([f], [g]) &\mapsto [f\circ g] .\end{align*}

The previous argument then precisely says:

• If $$u:B\to C$$ and $$f,g C\to D$$ with $$[f] = [g]$$, the composition $$[u] \circ [f] = [u] \circ [g]: B\to D$$ is well-defined, and

• If $$v:D\to E$$ and $$f,g: C\to D$$ with $$[f] = [g]$$, the composition $$[f] \circ [v] = [g] \circ [v]$$ is well-defined.

So this composition on $$\mathsf{K}$$ is well-defined. Moreover,

• There is an identity morphism $$[\one_A] \in \mathop{\mathrm{Hom}}_K(A, A)$$ coming from the class of $$\one_C \in \mathop{\mathrm{Hom}}_{\mathsf{Ch}( \mathcal{A} ) }(A, A)$$,

• It is associative, since $$[h] \circ [gf] \coloneqq[hgf] \coloneqq[hg] \circ [f]$$, and

• It is unital, in the sense that $$[\one_B] \circ [f] \coloneqq[\one_A \circ f] = [f] = [f \circ \one_A] \coloneqq[f] \circ [\one_B]$$ for any $$f:A\to B$$.

So this data satisfies all of the axioms of a category (Weibel A.1.1).

To see that $$f_0 + f_1 \simeq g_0 + g_1$$, we have \begin{align*} (f_0 + f_1) - (g_0 + g_1) &= (f_0 - g_0) + ( f_1 - g_1) \\ &\simeq 0 + 0 = 0 .\end{align*} To be explicit, if $$s_i$$ are chain homotopies for $$f_i, g_i$$, we can take $$s\coloneqq s_0 + s_1$$: \begin{align*} (f_0 + f_1) - (g_0 + g_1) &= (f_0 - g_0) + ( f_1 - g_1) \\ &= ({{\partial}}s_0 + s_0 {{\partial}}) + ({{\partial}}s_1 + s_1 {{\partial}}) \\ &= ({{\partial}}s_0 + {{\partial}}s_1) + (s_0 {{\partial}}+ s_1 {{\partial}}) \\ &= {{\partial}}(s_0 + s_1) + (s_0 + s_1){{\partial}}\\ &\coloneqq{{\partial}}s + s {{\partial}} .\end{align*}

That $$K$$ forms an additive category is a consequence of the following facts:

• $$K$$ has products, since $$\mathcal{A}$$ had products and $${\operatorname{Ob}}(K) \coloneqq{\operatorname{Ob}}( \mathcal{A})$$.

• $$K$$ has a zero object, for the same reason.

• Composition distributes over addition, i.e.  \begin{align*} [f] \qty{ [g] \oplus [h] } \\ &\coloneqq [f] [g +h] \\ &\coloneqq [f(g + h) ] \\ &\coloneqq [fg + fh] \\ &= [fg] \oplus [fh] .\end{align*}

Moreover, the quotient functor $$\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}}) \to K$$ is an additive functor, since the maps $$Q: \mathop{\mathrm{Hom}}_{\mathsf{Ch}( \mathcal{A} )}(A, B) \to \mathop{\mathrm{Hom}}_{\mathsf{Ch}( \mathcal{A} )}(A, B)/ \sim$$ are morphisms of abelian groups, using the fact that $$Q$$ commutes with both additions: \begin{align*} Q(f + g) = [f + g] \coloneqq[f] \oplus [g] = Q(f) + Q(g) .\end{align*} Alternatively, we can note that the set of all $$f\in \mathop{\mathrm{Hom}}_{ \mathsf{Ch}( \mathcal{A} }(A, B)$$ which are nullhomotopic form a subgroup $$H$$, and since everything is abelian we can form the quotient $$\mathop{\mathrm{Hom}}_{ \mathsf{Ch}(\mathcal{A}) } (A, B) / H$$ observe that this is isomorphic as a group to $$\mathop{\mathrm{Hom}}_K(A, B)$$.

We can first note that $$K$$ is an additive category, since we still have a zero object and products inherited from $$\mathsf{Ch}( \mathcal{A})$$.

Note: I don’t see a great way to prove that any particular category is abelian or not! Checking the axioms listed in Appendix A.4 seems quite difficult.

Let $$\operatorname{cone}(C) \coloneqq\operatorname{cone}(\one_C)$$, so \begin{align*} \operatorname{cone}(C)_n = C_{n-1} \oplus C_n .\end{align*} Show that $$\operatorname{cone}(C)$$ is split exact, with splitting map given by $$s(b, c) \coloneqq(-c, 0)$$.

Fixing notation, let

• $${{\partial}}_n$$ be the $$n$$th differential on $$C$$,

• $$\one_C$$ be the identity chain map on $$C$$,

• $$\one_n: C_n\to C_n$$ be the $$n$$th graded component of $$\one_C$$,

• $$\widehat{C} \coloneqq\operatorname{cone}(C) \coloneqq\operatorname{cone}(\one_C)$$,

• $$\widehat{{{\partial}}}_n$$ be the $$n$$th differential on $$\widehat{C}$$, and

• $$\widehat{\one}$$ be the identity on $$\widehat{C}$$,

• $$\widehat{\one}_n: \widehat{C}_n \to \widehat{C}_n$$ be the $$n$$th component of $$\widehat{\one}$$.

From exercise 1.4.2, it suffices to show that $$\widehat{\one}$$ is nullhomotopic. Since we have a direct sum decomposition $$\operatorname{cone}(C)_n \coloneqq C_{n-1} \oplus C_n$$, we can write $$\widehat{\one}$$ as a block matrix \begin{align*} \begin{bmatrix} \one_{n-1} & 0 \\ 0 & \one_n \end{bmatrix} .\end{align*}

We can similarly write down the differential on $$\operatorname{cone}(C)$$ in block form:

This yields \begin{align*} \widehat{{{\partial}}}_n \coloneqq \begin{bmatrix} -{{\partial}}_{n-1} & 0 \\ -\one_{n-1} & {{\partial}}_n \end{bmatrix} .\end{align*}

# Similarly, the map $$s_n(b, c) = (-c, 0)$$ can be written as  \begin{align*} s_n

\begin{bmatrix} 0 & -\one_n \ 0 & 0 \end{bmatrix} .\end{align*} {=html}

We can thus proceed by a direct computation: \begin{align*} {{\partial}}_{n+1} s_n + s_{n-1} {{\partial}}_n &= \begin{bmatrix} -{{\partial}}_n & 0 \\ -\one_n & {{\partial}}_{n+1} \end{bmatrix} \begin{bmatrix} 0 & -\one_n \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & -\one_{n-1} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} -{{\partial}}_{n-1} & 0 \\ -\one_{n-1} & {{\partial}}_n \end{bmatrix} \\ \\ &= \begin{bmatrix} 0 & {{\partial}}_n \one_n \\ 0 & \one_n \one_n \end{bmatrix} + \begin{bmatrix} \one_{n-1}\one_{n-1} & -\one_{n-1}{{\partial}}_n \\ 0 & 0 \end{bmatrix} \\ \\ &= \begin{bmatrix} \one_{n-1} \one_{n-1} & {{\partial}}_n \one_n - \one_{n-1} {{\partial}}_n \\ 0 & \one_n \one_n \end{bmatrix} \\ \\ &= \begin{bmatrix} \one_{n-1} \one_{n-1} & 0 \\ 0 & \one_n \one_n \end{bmatrix} \\ &= \widehat{\one}_n .\end{align*}

Let $$f:C\to D \in \mathop{\mathrm{Mor}}(\mathsf{Ch}(\mathcal{A}))$$ and show that $$f$$ is nullhomotopic if and only if $$f$$ lifts to a map \begin{align*} (-s, f): \operatorname{cone}(C) \to D .\end{align*}

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As a notational convention for this problem, I’ll take vectors $$v$$ to be column vectors, and $$v^t$$ will denote a row vector. I’ve also written $$f \simeq 0$$ to denote that $$f$$ is nullhomotopic.

$$\implies$$: Suppose that $$f \simeq 0$$, so there are maps $$s_n$$ such that the following diagrams commute for every $$n$$:

Write $$\widehat{C} \coloneqq\operatorname{cone}(C) \coloneqq\operatorname{cone}(\one_{C}) \coloneqq C[1] \oplus C$$, we then want to construct a lift $$\widehat{f}$$ of $$f$$ such that the following diagram commutes:

where $$\iota_{\widehat{C}}$$ is the following inclusion of $$C$$ into its cone:

We define the map $$\widehat{f}$$ in the following way:

That this is a lift follows from computing the composition, which can be done in block matrices:

#  \begin{align*} \widehat{f}n \circ \iota{\widehat{C}, n} &= \begin{bmatrix} -s_{n-1}\ f_n\end{bmatrix} ^t \begin{bmatrix} 0 \ \one_{C_n} \end{bmatrix}

## [ f_n \one_{C_n} ] = f_n ,\end{align*} {=html} where the first matrix acts as a row vector. It only remains to check that $\widehat{f}$ defines a chain map, which follows from the following computation:  \begin{align*} {{\partial}}_n^D \widehat{f}n - \widehat{f}{n-1} \widehat{{{\partial}}}_n &= [{{\partial}}n^D] \begin{bmatrix} -s{n-1} \ f_n\end{bmatrix} ^t

\begin{bmatrix} -s_{n-2} \ f_{n-1} \end{bmatrix} \begin{bmatrix} -{{\partial}}n^C & 0 \ -\one{C_n} & {{\partial}}{n+1}^C \end{bmatrix} \ \ &= \begin{bmatrix} {{\partial}}n^D (-s{n-1}) - s{n-2}{{\partial}}n^C + f{n-1} \one_{C_n} \ {{\partial}}n^D f_n - f{n-1} {{\partial}}n^C \end{bmatrix} ^t \ \ &= \begin{bmatrix} f{n-1} -( {{\partial}}n^D s{n-1} + s_{n-2}{{\partial}}_n^C) \ 0 \end{bmatrix} ^t && \text{since $$f$$ a chain map $$\implies {{\partial}}f=f {{\partial}}$$} \ \ &= \begin{bmatrix} 0 \ 0 \end{bmatrix} && \text{since $$f \simeq 0 \implies {{\partial}}s + s{{\partial}}= f$$} \ &=0 .\end{align*} {=html}

$$\impliedby$$: Suppose we have a lift $$\widehat{f}: \widehat{C} \to D$$; then define the following maps as the proposed splittings: \begin{align*} s_n: C_{n-1} &\to D_n \\ c &\mapsto \widehat{f}_n(-c, 0) .\end{align*}

There are two relevant facts to observe:

• We have $$f = \tilde f \iota_{\widehat{C}}$$ where $$\iota_{\widehat{C}}(c) \coloneqq(0, c) \in \widehat{C}$$ is inclusion into the second direct summand, and in particular \begin{align*} f_n(c) = \widehat{f}_n \iota_{\widehat{C}, n}(c) = \widehat{f}_n(0, c) .\end{align*}

• Since $$\widehat{f}$$ is a chain map, we have for each $$n$$ \begin{align*} {{\partial}}_n^D \widehat{f}_n(x, y) = \widehat{f}_{n+1} \widehat{{{\partial}}}_n(x, y) && \text{as maps } \widehat{C}_n \to D_n .\end{align*}

We now proceed to compute at the level of elements that $$s$$ defines a splitting:  \begin{align*} {{\partial}}{n+1}^D s_n(c) + s{n-1} {{\partial}}n^D(c) &\coloneqq {{\partial}}{n+1}^D \widehat{f}_{n+1}(c, 0) + \widehat{f}n {{\partial}}n^C (c, 0) \ \ &= {{\partial}}{n+1}^D \widehat{f}{n+1}(c, 0) + {\color{red} \widehat{f}_n ({{\partial}}n^C(c), 0) } \ \ &= {\color{red} \widehat{f_n} \widehat{{{\partial}}}{n+1} } (c, 0) + \widehat{f}n ({{\partial}}n^C(c), 0) && \text{by (2)} \ \ &= \widehat{f}n \qty{ \begin{bmatrix} -{{\partial}}{n}^C & 0 \ -\one{n}^C & {{\partial}}{n+1}^C \end{bmatrix} \begin{bmatrix} c
\ 0
\end{bmatrix} }

• \widehat{f}_n ({{\partial}}_n^C(c), 0) \ \ &= \widehat{f}_n \qty{ \begin{bmatrix} -{{\partial}}_n^C(c)
\ c \end{bmatrix} }
• \widehat{f}_n ({{\partial}}_n^C(c), 0) \ \ &= \widehat{f}_n (- {{\partial}}_n^C(c), c) + \widehat{f}_n( {{\partial}}_n^C(c), 0) \ &= \widehat{f}_n (- {{\partial}}_n^C(c) + {{\partial}}_n^C(c), c) && \text{since $$f_n$$ is an $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ morphism} \ &= \widehat{f}_n(0, c) \ &= f_n(c) && \text{by (1)} .\end{align*} {=html}
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• Show that free implies projective.

• Show that $$\mathop{\mathrm{Hom}}_R(M, {-})$$ is left-exact.

• Show that $$P$$ is projective if and only if $$\mathop{\mathrm{Hom}}_R(P, {-})$$ is exact.
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Suppose $$M$$ is free, so we have a set $$S$$ and an injection $$S \hookrightarrow M$$ such that every map in $$\hom_{{\mathsf{Set}}}(S, Y)$$ for $$Y\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ lifts to a unique map in $$\hom_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, Y)$$. Suppose further that we have the following situation; we seek to construct a lift $$\tilde h: M\to A$$:

This lift exists by first considering $$s_i \in S$$ and noting that since $$\beta_i \coloneqq\iota g(s_i) \in B$$ and $$f$$ is surjective, there exist some elements $$\alpha_i$$ in $$A$$ such that $$f(\alpha_i) = \beta_i$$ for each $$i$$. So define the map \begin{align*} h:S &\to A \\ s_i &\mapsto \alpha_i .\end{align*}

By the universal property of free modules, this lifts to a map $$\tilde h: M \to A$$, so $$M$$ is projective.

Suppose we have a SES \begin{align*} 0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0 .\end{align*} The claim is that $$\mathop{\mathrm{Hom}}_R(M, {-})$$ yields an exact sequence \begin{align*} 0 \to \mathop{\mathrm{Hom}}_R(M, A) \xrightarrow{f_*} \mathop{\mathrm{Hom}}_R(M, B) \xrightarrow{g_*} \mathop{\mathrm{Hom}}_R(M, C) \to \cdots ,\end{align*} where $$f_*(\alpha) = f \circ \alpha$$ for $$\alpha:M\to A$$ and similarly $$g_*(\beta) = g \circ \beta$$ for $$\beta:M\to B$$. To show that this is exact, it suffices to show three things:

• $$\ker f_* = 0$$,

• $$\operatorname{im}f_* \subseteq \ker g_*$$, and

• $$\ker g_* \subseteq \operatorname{im}f_*$$.

Proceeding with each part:

• By definition, if $$\beta: M\to B$$ satisfies $$\beta \in \ker f_*$$, and thus $$f\beta = 0$$. Since the original sequence was exact, $$f$$ is injective, thus a monomorphism, thus satisfies the left-cancellation property. So we can immediately conclude that $$\beta = 0$$.

• Let $$\beta: M\to B$$ be in $$\operatorname{im}f_*$$, so there exists some $$\alpha: M\to A$$ with $$\beta = f \alpha$$. We want to show that $$\beta\in \ker g_*$$, so we can apply $$g_*$$ to obtain $$g_*(\beta) = g_*(f \alpha) \coloneqq g f \alpha$$. But by exactness of the first sequence, $$gf = 0$$, so $$gfa = 0$$.

• Let $$\beta' \in \ker g_*$$, so $$g \beta' = 0$$. In order to show $$\beta' \in \operatorname{im}f_*$$, we want to construct some $$\alpha: M\to A$$ such that $$\beta' = f_*( \alpha) \coloneqq f \alpha$$. Considering $$m\in M$$, we know $$g \beta'(m) =0$$ and thus $$\beta'(m) \in \ker g = \operatorname{im}f$$ by exactness of the first sequence. So there exists some $$a_m\in A$$ with $$f(a_m) = \beta'(m)$$, and we can define a map \begin{align*} \alpha: M\to A \\ m & \mapsto a_m .\end{align*} By construction, we then have $$f \alpha(m) = f(a_m) \coloneqq\beta'(m)$$ for every $$m\in M$$, so $$f \alpha = \beta'$$.

Assume the same setup as (b).

$$\implies$$: Suppose $$P$$ is projective, so it satisfies the following universal property:

Using the results of (b), it suffices to check exactness at $$\mathop{\mathrm{Hom}}_R(P, C)$$ in the following sequence: \begin{align*} 0 \xrightarrow{} \mathop{\mathrm{Hom}}_R(P, A) \xrightarrow{f_*} \mathop{\mathrm{Hom}}_R(P, B) \xrightarrow{g_*} \mathop{\mathrm{Hom}}_R(P, C) \to 0 ,\end{align*} or equivalently that $$g_*$$ is surjective. Using that $$B \xrightarrow{g} C\to 0$$ is exact if and only if $$g$$ is surjective, the universal property above means that every $$\gamma \in \mathop{\mathrm{Hom}}_R(P, C)$$ lifts to a map $$\tilde \gamma \in \mathop{\mathrm{Hom}}_R(P, B)$$ where $$g\tilde \gamma = \gamma$$. Since $$g_*(\tilde \gamma) \coloneqq g \tilde \gamma$$, this precisely means that $$\gamma \in \operatorname{im}g_*$$.

$$\impliedby$$: Reversing the above argument, if $$\mathop{\mathrm{Hom}}_R(P, {-})$$ is exact, then every $$P \xrightarrow{\gamma} C$$ has a preimage under $$g_*$$, which is precisely a lift $$P \xrightarrow{\tilde \gamma} B$$. So $$P$$ satisfies the universal property of projective modules.