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3. Let f : S → S ′ be any map of sets. Applying in the diagram<br />
S<br />
ι S<br />
f<br />
<br />
S ′ ι S ′<br />
V (S)<br />
<br />
V (f)<br />
<br />
<br />
V (S ′ )<br />
the defining property of the free vector space V (S) to the map ι S ′ ◦ f of sets, we find a<br />
unique K-linear map V (f) : V (S) → V (S ′ ).<br />
One checks that this defines a functor V : Set → vect(K). For any K-vector space W and<br />
any set S, we have a bijection of morphism spaces:<br />
Hom K (V (S), W ) ∼ = Hom Set (S, U(W ))<br />
that is compatible with morphism of sets and K-linear maps. One says that the functor<br />
V is a left adjoint to the forgetful functor U.<br />
A.2 Tensor products of vector spaces<br />
We summarize some facts about tensor products of vector spaces over a field K.<br />
Definition A.2.1<br />
Let K be a field and let V, W and X be K-vector spaces. A K-bilinear map is a map<br />
α : V × W → X<br />
that is K-linear in both arguments, i.e. α(λv + λ ′ v ′ , w) = λα(v, w) + λ ′ α(v ′ , w) and α(v, λw +<br />
λ ′ w ′ ) = λα(v, w) + λ ′ α(v, w ′ ) for all λ, λ ′ ∈ K and v, v ′ ∈ V , w, w ′ ∈ W .<br />
Given any K-linear map ϕ : X → X ′ , the map ϕ ◦ α : V × W → X ′ is K-bilinear as well.<br />
This raises the question of whether for two given K-vector spaces V, W , there is a “universal”<br />
K-vector space with a universal bilinear map such that all bilinear maps out of V × W can be<br />
described in terms of linear maps out of this vector space.<br />
Definition A.2.2<br />
The tensor product of two K-vector spaces V, W is a pair, consisting of a K-vector space V ⊗W<br />
and a bilinear map<br />
κ : V × W → V ⊗ W<br />
(v, w) ↦→ v ⊗ w<br />
with the following universal property: for any K-bilinear map<br />
α : V × W → U<br />
there exists a unique linear map ˜α : V ⊗ W → U such that<br />
α = ˜α ◦ κ .<br />
As a diagram:<br />
V × W<br />
<br />
<br />
U<br />
<br />
<br />
<br />
κ<br />
∃!˜α<br />
V ⊗ W<br />
α<br />
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