14 SERGE BOUC

by objects of V, such that there is a commutative diagram

0

0

QM

I

PM

I

M

—

QN

VN

I

'•*

QL

i

- P'N £

PL

VNI

- • N

--

i

» L

0

0

with exact rows and columns (note that this resolution needs not be equal to the

prescribed resolution of N).

Then it is easy to check that the following diagram is commutative

F

F'

0

F(PM®P'N)

0

F'(M®N)

I

0

F(Q'N

AF{a)

7 0

0 0

N)

F(P'N

F'WN)

AF'(^)

F'(N)

i

0

F(P)

F'W

F(Qi

AF(pL)

F(PL)

F'(1L)

F'(L)

I

0

Moreover, its columns are exact by the above remarks, as well as its two top lines.

Now the bottom line is also exact: the map F'{tp) o F'{ij)'N) = F{ip'L) o F((3) is an

epimorphism, because F(ip'L) and F{(3) are. Hence F'(ip) is also an epimorphism.

And if 9:F'{N)-B is any map in B such that 0 o AF'{ip) = 0, then

6 o A F ' M o F' ( ^ £

N

) = 0 ° F'WN) ° AF(a) = 0

As the middle row is exact, there is a map \i : F{PL) — • B such that

6oF'(ip'N)=tioF(f3)

Now

0 = 0 o F'WN) o AF(p'N) = \x o F{p) o AF(p'N) = /x o

AF&L)

o f ( j ° )

As the top line is exact, I have

fioAF(ifL)=0

B such that JJL = and as the right column is exact, there is a map A : F'(L)

A O F ' C 0 L ) .

Thus

0 o F'WN) = A o

F'WL) O

F(f3) = A o F'ty) o F'^N)

As Ff(ipfN) is an epimorphism, I have 9 = A oF'(?/), so the bottom line of the above

diagram is exact, and the functor F' is right exact. This completes the proof of the

theorem. •