Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.