Quasi-isomorphism of injective complexes is a homotopy equivalence?

Question

Let $A^* = (0\rightarrow A^0\rightarrow A^1 \rightarrow ...), B^* = (0\rightarrow B^0\rightarrow B^1 \rightarrow ...)$ be complexes of injective objects of an abelian category $\mathcal{A}$. Suppose we have a quasi-isomorphism $f : A^* \rightarrow B^*$, i.e. the induced cohomology map $f^*:H^*(A^*) \rightarrow H^*(B^*)$ is an isomorphism. How do I show that $A^*$ and $B^*$ are homotopic, i.e. there exists $g: B^*\rightarrow A^*$ such that $gf \simeq 1_A$ and $fg \simeq 1_B$?

My attempt: So I tried to define $g: B^*\rightarrow A^*$ from the inverse $g^*: H^*(B^*)\rightarrow H^*(A^*)$ of $f^*$. Using \begin{equation} \begin{array} 00 & \rightarrow & H^0(B^*) & \rightarrow & B^0\\ & & \downarrow & & \\ 0 & \rightarrow & H^0(A^*) & \rightarrow & A^0 \\ \end{array} \end{equation} I have that $g^0: B^0 \rightarrow A^0$ exists because $A^0$ is injective. Then I would like to say that if $g^0,g^1,...,g^{i-1}$ exist then $g^i$ must also exists: \begin{equation} \begin{array} 00 & \rightarrow & B^0 & \rightarrow & B^1 & \rightarrow & ... & \rightarrow & B^{i-1} &\rightarrow & B^i\\ & & \downarrow & & \downarrow & & & & \downarrow & & \\ 0 & \rightarrow & A^0 & \rightarrow & A^1 & \rightarrow & ... & \rightarrow & A^{i-1} &\rightarrow & A^i \\ \end{array} \end{equation} The problem is that $B^{i-1}\rightarrow B^i$ is not a monomorphism and I'm not sure how to continue from here.

Answer 1

We consider the mapping cone $C(f)^i := A^{i+1}\oplus B^i$. The differentials of $C(f)$ is \begin{equation} D^i = \begin{pmatrix} -d^{i+1}_A & 0\\ f^{i+1} & d^i_B \end{pmatrix} \end{equation} Since $f^*:H^i(A^*)\rightarrow H^i(B^*)$ is an isomorphism for all $i$, by taking the long exact sequence of cohomology of \begin{equation} 0\rightarrow B^* \rightarrow C(f)\rightarrow A[1]^*\rightarrow 0 \end{equation} we find that $H^i(C(f)^*) = 0, \forall i$, i.e. $C(f)$ is quasi-isomorphic to zero. But $A^i$ and $B^i$ are injective objects, so $C(f)^i$ is also injective as a finite coproduct (which is the same as product) of injective objects. So $C(f)$ is an injective resolution of $0$, just like the zero complex $0\rightarrow 0\rightarrow ...$ hence they are homotopic, i.e. $1_{C(f)} \simeq 0_{C(f)}$. This is equivalent to \begin{equation} 1_{C(f)} = D^{i-1}\Sigma^i + \Sigma^{i+1}D^i \end{equation} for some morphism $\Sigma^*$. We can write $\Sigma^*$ in general as \begin{equation} \Sigma^i = \begin{pmatrix} \Sigma_{AA}^i & \Sigma_{AB}^i\\ \Sigma^i_{BA} & \Sigma^i_{BB} \end{pmatrix} \end{equation} where $\Sigma^i_{AA}: A^{i+1}\rightarrow A^i,\ \Sigma^i_{AB}:B^i\rightarrow A^i,\ \Sigma^i_{BA}: A^{i+1}\rightarrow B^{i-1},\ \Sigma^i_{BB}:B^i\rightarrow B^{i-1}$. Let $(a^{i+1},b^i)\in C(f)^i$. Then $(a^{i+1},b^i) = 1^i_{C(f)}(a^{i+1},b^i) = D^{i-1}\Sigma^i(a^{i+1},b^i) + \Sigma^{i+1}D^i(a^{i+1},b^i)$ gives \begin{align} a^{i+1} &= -d_A^i\Sigma_{AA}^ia^{i+1} - d_A^i\Sigma^i_{AB}b^i - \Sigma^{i+1}_{AA}d^{i+1}_Aa^{i+1} + \Sigma_{AB}^{i+1}f^{i+1}a^{i+1} + \Sigma_{AB}^{i+1}d^i_Bb^i\\ b^i &= f^i\Sigma_{AA}^ia^{i+1} + f^i\Sigma_{AB}^ib^i + d^{i-1}_B\Sigma^i_{BA}a^{i+1}+d_B^{i-1}\Sigma_{BB}^ib^i - \Sigma_{BA}^{i+1}d_A^{i+1}a^{i+1} + \Sigma_{BB}^{i+1}f^{i+1}a^{i+1} + \Sigma_{BB}^{i+1}d_B^ib^i. \end{align} Since $(a,b)$ is arbitrary, by setting $b = 0$ we get \begin{align} a^{i+1} - \Sigma_{AB}^{i+1}f^{i+1}a^{i+1} &= -d_A^i\Sigma_{AA}^ia^{i+1} - \Sigma^{i+1}_{AA}d^{i+1}_Aa^{i+1}\tag{1}\\ \Sigma_{BA}^{i+1}d_A^{i+1}a^{i+1} - d^{i-1}_B\Sigma^i_{BA}a^{i+1} &= f^i\Sigma_{AA}^ia^{i+1} + \Sigma_{BB}^{i+1}f^{i+1}a^{i+1}\tag{2}. \end{align} On the other hand, by setting $a = 0$ we get \begin{align} d_A^i\Sigma^i_{AB}b^i &= \Sigma_{AB}^{i+1}d^i_Bb^i\tag{3}\\ b^i - f^i\Sigma_{AB}^ib^i &= d_B^{i-1}\Sigma_{BB}^ib^i + \Sigma_{BB}^{i+1}d_B^ib^i\tag{4}. \end{align} Equation (3) shows that $g := \Sigma_{AB}^*:B^*\rightarrow A^*$ is a morphism of complexes. Equation (1) and (4) shows that $gf\simeq 1_A$ and $fg \simeq 1_B$ respectively. Equation (2) seems to give an extra identity, but I don't think it is needed here.

Answer 2

This is a direct consequence of Lemma 10.4.6 in Weibel's Introduction to Homological Algebra which roughly says

If $A$ is a bounded below chain complex of injectives and $f \colon A \to B$ is a quasi-isomorphism then $f$ is left invertible in the homotopy category $K(\mathcal{A})$.

In your situation both $A$ and $B$ are bounded below complexes of injectives, so you can apply the lemma twice: In $K(\mathcal{A})$ the morphism $f$ is left invertible, and any left inverse $g$ of $f$ is a quasi-isomorphism and hence left invertible again. Now clearly (in any category) a morphism which has a left invertible left inverse is an isomorphism.

Sketch of a proof of Lemma 10.4.6: We may assume without loss of generality that $B$ is bounded below as well, say $A^i = B^i = 0$ for all $i \leq 1$. There is a canonical morphism $\varphi \colon C(f) \to A[1]$, where $C(f)$ denotes the mapping cone of $f$. Since $f$ is a quasi-isomorphism, $C(f)$ is exact, and so can be regarded as a resolution of $C(f)^0 = 0$. As a morphism between a resolution and a chain complex of injectives, $\varphi$ is uniquely determined up to homotopy by $\varphi^0$ which is the zero map. Hence $\varphi$ is null-homotopic, say by a homotopy $(t_i,g_i) \colon A^{i+1} \oplus B^i \to A^i$. Now you can check that $g$ is a morphism $B \to A$ and that $g \circ f$ is equivalent to the identity map witnessed by the homotopy $t$.