Tensor product of $K$ -linear maps

Suppose $𝑓 : 𝑉 1 \to 𝑉 2$ and $𝑔 : 𝑊 1 \to 𝑊 2$ are $K$ -linear maps. Then there is a unique $K$ -linear map $𝑓 \otimes 𝑔 : 𝑉 1 \otimes 𝑊 1 \to 𝑉 2 \otimes 𝑊 2$ , called the tensor product of the maps $𝑓$ and $𝑔$ , such that

c|https://q.uiver.app/#q=WzAsNCxbMiwwLCJWXzEgXFxvdGltZXMgV18xIl0sWzIsMiwiVl8yIFxcb3RpbWVzIFdfMiJdLFswLDIsIlZfMiBcXHRpbWVzIFdfMiJdLFswLDAsIlZfMSBcXHRpbWVzIFdfMSJdLFswLDEsImYgXFxvdGltZXMgZyJdLFsyLDEsIihcXG90aW1lcykiLDJdLFszLDAsIihcXG90aW1lcykiXSxbMywyLCJmIFxcdGltZXMgZyIsMl1d

commutes. #m/thm/module

Proof

We claim that the map

ℎ : = (\otimes) \circ (𝑓 \times 𝑔) : 𝑉 1 \times 𝑊 1 \to 𝑉 2 \otimes 𝑊 1

along the diagonal is bilinear, the conclusion then follows from the Universal property of the tensor product. Let $𝑣 1, 𝑣 2 \in 𝑉 1$ , $𝑤 \in 𝑊 1$ , and $𝛼, 𝛽 \in K$ . Then

ℎ (𝛼 𝑣 1 + 𝛽 𝑣 2, 𝑤) = 𝑓 (𝛼 𝑣 1 + 𝛽 𝑣 2) \otimes 𝑔 𝑤 = (𝛼 𝑓 𝑣 1 + 𝛽 𝑓 𝑣 2) \otimes 𝑔 𝑤 = 𝛼 𝑓 𝑣 1 \otimes 𝑔 𝑤 + 𝛽 𝑓 𝑣 2 \otimes 𝑔 𝑤 = 𝛼 ℎ (𝑣 1, 𝑤) + 𝛽 ℎ (𝑣 2, 𝑤)

so $ℎ$ is linear when the second argument is held constant. A similar calculation holds for the first argument held constant. Therefore $ℎ$ is bilinear, as claimed.

This makes $(\otimes K) : K 𝖬 𝗈 𝖽 \times K 𝖬 𝗈 𝖽 \to K 𝖬 𝗈 𝖽$ into a bifunctor and $K 𝖬 𝗈 𝖽$ into a monoidal category.

Tensor product of K-linear maps

See also

Tensor product of $K$ -linear maps