$𝑅$ -tensor product of linear maps

Suppose $𝑓 \in 𝖬 𝗈 𝖽 𝑅 (𝑉 1, 𝑉 2)$ and $𝑔 \in 𝑅 𝖬 𝗈 𝖽 (𝑊 1, 𝑊 2)$ . Then there is a unique morphism $𝑓 \otimes 𝑔 : 𝖠 𝖻 (𝑉 1 \otimes 𝑊 1, 𝑉 2 \otimes 𝑊 2)$ , 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 balanced, the conclusion then follows from the Universal property of the tensor product. Let $𝑣, 𝑣' \in 𝑉 1$ , $𝑤, 𝑤' \in 𝑊 1$ , and $𝛼 \in 𝑅$ . First, we have

ℎ (𝑣, 𝑤 + 𝑤') = 𝑓 𝑣 \otimes 𝑔 (𝑤 + 𝑣 𝑤) = 𝑓 𝑣 \otimes (𝑔 𝑤 + 𝑔 𝑤') = 𝑓 𝑣 \otimes 𝑔 𝑤 + 𝑓 𝑣 \otimes 𝑔 𝑤' = ℎ (𝑣, 𝑤) + ℎ (𝑣, 𝑤')

giving ^B2. Similarly

ℎ (𝑣 + 𝑣', 𝑤) = 𝑓 (𝑣 + 𝑣') \otimes 𝑔 𝑤 = (𝑓 𝑣 \otimes 𝑓 𝑣') \otimes 𝑔 𝑤 = (𝑓 𝑣 + 𝑓 𝑣') \otimes 𝑔 𝑤 = 𝑓 𝑣 \otimes 𝑔 𝑤 + 𝑓 𝑣' \otimes 𝑔 𝑤 = ℎ (𝑣, 𝑤) + ℎ (𝑣', 𝑤)

giving ^B2. Finally,

ℎ (𝑣 𝛼, 𝑤) = 𝑓 (𝑣 𝛼) \otimes 𝑔 𝑤 = (𝑓 𝑣) 𝛼 \otimes 𝑔 𝑤 = 𝑓 𝑣 \otimes 𝛼 𝑔 𝑤 = 𝑓 𝑣 \otimes 𝑔 (𝛼 𝑤) = ℎ (𝑣, 𝛼 𝑤)

giving ^B3. Therefore $ℎ$ is balanced, as claimed.

𝑅-tensor product of linear maps

See also

$𝑅$ -tensor product of linear maps