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Say that we have $R$-modules (let's assume over a commutative ring $R$). Consider extensions of a module $A$ by a module $C$, so that we have short exact sequences: $0 \rightarrow B \rightarrow^i E \rightarrow C \rightarrow 0.$ We say that two extensions are equivalent if they can be fitted into a commutative diagram:

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ aextensions

Now, if we have two $R$-module homomorphisms $f\colon B\to C$, $g\colon B\to C$, we can from f and g indivudally create new extensions, where the middle term is the pushout of $f$ and $i$, resp. $g$ and $i$.

When will these two homomorphisms determine the same extension, that is, when will we be given an equivalence of extensions (as in the definition before this)?

I'm working through a problem set that says that this happens if and only if $f-g$ can be written as $i\colon B\to E$, followed by a homomorphism $E\to C$, but I can't see why this should be true, if the middle term E is not projective.

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1 Answer 1

up vote 3 down vote accepted

You don't need projectivity of $E$ because you have sufficiently many push-out diagrams and short exact sequences around.


Let us fix the notation first. Form the push-out under $i$ and $f$ to obtain the diagram

    i       i'
B >---> E --->> A
|       |       |
|f      |f'     |1
V   j   V   j'  v
C >---> P --->> A

and form the push-out under $i$ and $g$ to obtain the diagram

    i       i'
B >---> E --->> A
|       |       |
|g      |g'     |1
V   k   V   k'  v
C >---> Q --->> A.

We want to show that the extensions C >-> P ->> A and C >-> Q ->> A are equivalent if and only if there exists a morphism m: E -> C such that mi = g - f.


To say that C >-> P ->> A and C >-> Q ->> A are equivalent is to say that there is an isomorphism l: P -> Q such that

    j       j'  
C >---> P --->> A
|       |       |
|1      |l      |1
V   k   V   k'  v
C >---> Q --->> A

commutes. Consider the morphism g'-lf': E -> Q. We have

k'(g'-lf') = k'g' - k'lf' = i' - j'f' = 0

so that (g'-lf') = km for a unique morphism m: E -> C since k = ker k'. Now notice that

kmi = (g' - lf')i = g'i - lf'i = kg - ljf = k (g - f)

so that the fact that k is a monomorphism gives that mi = g-f, as we wanted.


Conversely, suppose there is a morphism m: E -> C such that mi = g-f. Then

kf = k (g - mi) = kg - kmi = g'i - kmi = (g' - km) i

so that we have a commutative square

    i    
B >---> E
|       |
|f      |g'-km
V   k   V
C >---> Q

Applying the push-out property of BECP to that square we get a unique morphism l: P -> Q such that k = lj and g'-km = lf' and I claim that the diagram

    j       j'  
C >---> P --->> A
|       |       |
|1      |l      |1
V   k   V   k'  v
C >---> Q --->> A

commutes. For the left hand square this is given and for the right hand square notice that

(k'l - j') f' = k'lf' - j'f' = k' (g' - km) - i' = k'g' - i' = 0

and

(k'l - j') j = k'lj = k'k = 0

so that (k'l - j')[j f'] = [0 0] and this implies that k'l-j' = 0 because the morphism [j f']: C+E -->> P is an epimorphism, as BECP is a push-out by hypothesis. It follows from the five lemma that the two extensions C >-> P ->> A and C >-> Q ->> B are equivalent.


It is a good exercise to show that a similar argument applied to an appropriate square (I'll let you find the morphism (?) yourself)

    i
B >---> E
|       |
|g      |(?)
V   j   V
C >---> P

and exploiting the push-out property of BECQ yields a morphism l': Q -> P which is inverse to l: P -> Q.

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