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is exact. Using the universal property of the direct sum, this amounts to
0 → Hom K (M, T ′ ) × Hom K (N, T ′ ) → Hom K (M, T ) × Hom K (N, T )
→ Hom K (M, T ′′ ) × Hom K (N, T ′′ ) → 0 .
The kernel of a Cartesian product of maps is the product of kernels; the image of the
Cartesian product of maps is the Cartesian product of the images. This implies the exactness
of the sequence in 4.
4⇒ 1 From the surjectivity of the line, we get a short exact sequence
By 4, we get a short exact sequence
0 → ker((N 1 → N 2 )) → N 1
f
→ N2 → 0
0 → Hom R (M, ker(N 1 → N 2 )) → Hom R (M, N 1 ) → Hom R (M, N 2 ) → 0
where the last arrow is
f ∗ : Hom R (M, N 1 ) → Hom R (M, N 2 )
ϕ ↦→ f ◦ ϕ =: f ∗ (ϕ)
The surjectivity of this morphism amounts to property 1.
✷
Definition 3.2.6
An A-module with one of the four equivalent properties from proposition 3.2.5 is called a
projective module.
Proposition 3.2.7.
Let A be a K-algebra. Then the following assertions are equivalent:
1. The algebra A is semisimple, i.e. seen as a left module over itself, it is a direct sum of
simple submodules.
2. Any A-module is semisimple, i.e. direct sum of simple submodules.
3. The category A−mod is semisimple, i.e. all A-modules are projective.
As a consequence of this result, we need to understand only simple modules to understand
the representation category of a semisimple algebra.
Proof.
3.⇒ 2. Suppose that the category A−mod is semisimple. Let M be an A-module. Any submodule
U ⊂ M yields a short exact sequence
0 → U → M → M/U → 0 ,
which splits, since the module M/U is projective. Then M ∼ = U ⊕M/U and the submodule
U has a complement.
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