# On injective resolutions

Let $h:M \rightarrow N$ be a module homomorphism and $((F,e);\alpha)$ be a right resolution of $M$ and $((I,d);\beta)$ be an injective resolution of $N$. If $f,g:(F,e) \rightarrow (I,d)$ are morphisms of cocoplexes, then I know that $f,g$ are homotopic. How do we construct a homotopy?

-
Read the proof of the comparison theorem for projective resolutions e.g. in Weibel (or any other book on homological algebra). Then dualize. –  commenter Nov 22 '12 at 13:38
So, I tried, but I can't construct homotopy.... –  Sang Cheol Lee Nov 22 '12 at 16:30

Depending on your notation you probably should have mentioned that $f$ and $g$ are maps that extend $h$.
The proof is the same as in the projective case. Form the chain map $k = f - g$. Since $k\alpha = 0$, we can quotient to get a diagram $I_0 \leftarrow F_0/\mathrm{im}(a)\hookrightarrow F_1$. The injective property gives us the $s_1:F_1\to I_0$. To continue the construction, consider the map $k_1 - ds$; I'll leave it as an exercise to show that $(k_1-ds)\circ d = 0$ (perhaps this step was the point of confusion?). At this point the rest of the proof should be clear (once you draw the diagrams, of course); this should be enough for you to finish the construction by induction.