Math 811, Ping Li

Exercise 7A: a) Prove that in an abelian category a morphism which is both mono and epi is an isomorphism.

b) Show, with an example, that this is not necessarily true in an arbitrary category. Remark: By the Freyd-mitchell embedding theorem, we may assume that the

abelian category is mod-R. That it is well known that a mono and epi is an isomorphism. However, one can also give a direct category, and the detailed argument is as follows:

Proof: a) Letf : A → B be mono and epi morphism in a abelian category. To prove f is an isomorphism, we must show that there is unique map g : B → A, which satisfies fg = 1_{B} and gf = 1_{A}.

we have known that the cokernel of f is the natural map.

π : B → B/Im(f) = 0

( because f is epi, im(f) = B) and the kernel of f is the inclusion map

ι : 0 = ker(f) → A

(Since f is mono). So f is the kernel of π, and the cokernel of ι.

f is the kernel of π, so let 1_{B} : B → B, π1_{B} = 0, then there is a unique morphism g : B → A, such that fg = 1_{B}.

\xymatrix{B \ar@{.>}[d]_{g} \ar[dr]^{1_{B}} \\ A \ar[r]^{f} & B \ar[r]^{π} & 0}

Let 1_{A} : A → A,1_{A}ι = 0, then there is a unique morhism g′ : B → A, such thatg′f = 1_{A}.

\xymatrix{0 \ar[r]^{ι} & A \ar[r]^{f} \ar[dr]_{1_{A}} & B \ar@{.>}[d]^{g′} \\  &  & A}

Next we have to show g = g′.

Since

fg′f = f(g′f) = f = (fg)f = fgf

and f is epi and mono, then g′ = g.

Thus g is the required map which satisfied fg = 1_{B}, gf = 1_{A}. so f is an isomorphism.

b)According to Exercise 6D in Tony’s notes, the inclusion of Z in Q, i.e.,

i : Z → Q

is mono and epi, but i is not an isormorphism in Ring category.

Sketch of proof: It is easy to see that i is mono. Since i is an inclusion. So, it is suffices to prove that i is epi.

Consider morphisms of Ring category f, g : Q → R. Suppose fi = gi, we want to show that f = g.

For all a ∈ Z,

f(a) = fi(a) = gi(a) = g(a)

and

f(1) = f(aa^{−1}) = f(a)f(a^{−1}) = 1

i.e.

f(a^{−1}) = f(a)^{−1}

similarly

g(a^{−1}) = g(a)^{−1}

Thus ∀a/b ∈ Q, b ≠ 0

f(\frac{a}{b}) = f(ab^{−1}) = f(a)f(b)^{−1} = \frac{f(a)}{f(b)} = \frac{g(a)}{g(b)} = g(\frac{a}{b})

then f = g. So i is epi.