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 $\mathcal N$ be an exact category and $C\mathcal N$ be the category of chain complexes with its usual exact structure. We have here the usual notion of "mapping cylinder" of a chain map. If $f:N\to M$ is a chain map, denote by $T(f)$ its mapping cylinder. Denote by $j_1:N\to T(f)$ and $j_2:M\to T(f)$ the inclusion on the first and third factors respectively. These are admissible monomorphisms.

Suppose we have a commutative diagram in $C\mathcal N$:

enter image description here

where $\alpha$ and $\beta$ are admissible monomorphisms.

Let $t:T(f)\to T(f')$ be the induced map; that is, $t=\left(\begin{smallmatrix} \alpha & 0 & 0 \\ 0 & \alpha[-1] & 0 \\ 0 & 0 & \beta \end{smallmatrix}\right)$.

I want to prove that the induced arrow $s$ in the pushout diagram below is an admissible monomorphism:

enter image description here

Notice that every arrow there but $s$ is an admissible monomorphism.

So I was able to show that $s$ has a cokernel (it's the cokernel $e$ of the map $T(f)\oplus (A'\oplus B') \to T(f')$ given by the matrix $(t \hspace{.6cm} j_1'\oplus j_2')$).

But I'm stuck at showing that $s$ is also the kernel of $e$. I've tried several different things for a couple of days now but I'm just absolutely stuck...

share|cite|improve this question
Oh, I wish @t.b. wasn't "on extended leave" from the site... – Bruno Stonek May 16 '14 at 17:17
up vote 1 down vote accepted

It's probably easiest to compute $P$ and the map $s$ explicitly. It turns out that $P$ has components $P_n = A_{n}' \oplus A_{n-1} \oplus B_{n}'$ and that $s_n = 1_{A_{n}'} \oplus \alpha_{n-1} \oplus 1_{B_{n}'}$, which is obviously an admissible monomorphism since it is the direct sum of admissible monomorphisms.

Some details, using Weibel's sign conventions on p.20:$\require{AMScd}$

The complex $T(f)$ has components $A_{n} \oplus A_{n-1} \oplus B_n$ and differential $\begin{bmatrix} d^{A}_{n} & 1_{A_{n-1}} & 0 \\ 0 & -d^{A}_{n-1} & 0 \\ 0 & -f_{n-1} & d^{B}_{n} \end{bmatrix}$, similarly for $T(f')$.

In each degree $n$ there is the push-out diagram $$\begin{CD} A_n \oplus B_n @>{\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1\end{bmatrix}}>> A_n \oplus A_{n-1}\oplus B_n @>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>> A_{n-1} \\ @VV{\begin{bmatrix} \alpha_n & 0 \\ 0 & \beta_n\end{bmatrix}}V @VV{\begin{bmatrix} \alpha_n & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \beta_n \end{bmatrix}}=k_nV @| \\ {A_{n}' \oplus B_{n}'} @>>{\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1\end{bmatrix}}=q_n> {A_{n}' \oplus A_{n-1} \oplus B_{n}'} @>>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}> A_{n-1} \end{CD}$$ from which you can see that $P_n = A_{n}' \oplus A_{n-1} \oplus B_{n}'$. Contemplating the push-out above, you can convince yourself that the differential of $P$ must be given by the matrix $$ \begin{bmatrix} d_{n}^{A'} & \alpha_{n-1} & 0 \\ 0 & - d_{n-1}^A & 0 \\ 0 & -\beta_{n-1} f_{n-1} & d_{n}^{B'} \end{bmatrix} $$ and that the induced map $s_n\colon P_n \to T(f')$ is given by $s_n = 1_{A_{n}'} \oplus \alpha_{n-1} \oplus 1_{B_{n}'}$, as desired.

Note also that this shows that the cokernel of $s$ is equal to the cokernel of $\alpha[-1]$.

share|cite|improve this answer
Great! In fact, not too many days ago I managed to solve the problem with help from my supervisor, and I was planning to post it here very soon. I got the same thing as you though I had to use elements to be convinced that the $P$ you define is effectively the pushout, etc. Thanks a lot for your effort on an already old and quite ignored question. – Bruno Stonek Jun 17 '14 at 17:56
For the prospective reader, I might add that the easy observation that the forgetful functor from $C\mathcal N$ to the category of graded objects over $\mathcal N$ reflects exactness is implicitly being used. – Bruno Stonek Jun 20 '14 at 17:13

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.