Non trivial extensions of the zero map

Question

Suppose we have short exact sequences of $R$-modules $0 \to A \to B \to C \to 0$ and $0 \to A' \to B' \to C' \to 0$ with a map of short exact sequences $f_A : A \to A', f_B : B \to B', f_C : C \to C'$.

If $f_A =0, f_C = 0$, is it true that $f_B = 0$? What if both short exact sequences are split?

Answer 1

1. It's not clear to me what saying "$f_A : A \to A'$ is a short exact sequence" means (it's been a long time since I did this stuff). I think that any sequence of two R-modules is "exact" because at each module, you have only one map. Did you perhaps mean that $0 \to A \to A' \to 0$ is short exact? If so, then $f_A$ is an isomorphism.

2. As written, if I'm following what you've asked...

Consider

0 -> Z  -> Z -> Z/2Z -> 0
     |     |      |
     V     V      V
0 -> Z  -> Z -> Z/2Z -> 0

where the first horizontal map in each row is $x \to 2x$, and the first and third vertical maps are zero, but the middle map is the identity.

This has all the properties you specified, but the middle vertical map isn't zero.

I believe you wanted to add the assumption that the diagram consisting of all your maps is actually commutative. Even with that assumption, I'm not sure of whether the statement is true (although I suspect it might be).

Answer 2

It's false. Suppose $A=C$ and $B= A \oplus A$. The map $f_B = A \oplus A \to A \oplus A$ given by $(a,b) \mapsto (b,0)$ is zero when restricted to the first component, and is zero if we quotient out the first component, but it's not zero globally.

Answer 3

The answer is "no" even if both are split, but this is not strange if you are used to nilpotent maps: suppose that $R=\mathbb R$ say, and consider the short exact sequence

$$ 0 \to \mathbb R \to^i \mathbb R^2 \to^q \mathbb R \to 0 $$ where the first map is given by $i(t)=te_1$ and the second map is given by $q(ae_1+be_2)=b$.

This short exact sequence essentially just filters $\mathbb R^2$ by the line $L=\mathbb R.e_1$, so that a map $\alpha\colon \mathbb R^2 \to \mathbb R^2$ induces an endomorphism of the short exact sequence provided it preserves $L$ i.e. its matrix $A$ with respect to the standard basis is upper triangular, so $A = \left(\begin{array}{cc}a & b \\ 0 & c \end{array}\right) = aE_{11}+bE_{12} +cE_{22}$ -- indeed that is exactly what is required in order for $\alpha$ to induce maps between the two outer terms in the short exact sequence. Those maps on the outer terms, however, are given by the diagonal terms $aE_{11}$ and $cE_{22}$ respectively, and so requiring that they are zero places no restriction on $bE_{12}\in \text{Hom}(\mathbb R^2,i(\mathbb R))$.