universal property of tensor product-the canonical map

Question

Consider tensor products in modules. Suppose $$f:M\times (N\oplus L)\to (M\otimes N)\oplus (M\otimes L)$$ is that alleged canonical map. How do I define $$g:M\times (N\oplus L)\to K$$ with the property that there exists a unique $$h:(M\otimes N)\oplus (M\otimes L)\to K$$ with $$h\circ \hat{f}=\hat{g}$$ ? By all this I want prove that $${\hat{f}}:M\otimes (N\oplus L)\to (M\otimes N)\oplus (M\otimes L)$$ is an isomorphism.

Answer 1

I am not sure if you can use the universal property of the tensor product to show that you can get a map that factors through $(M \otimes N) \oplus (M \otimes L)$, as you're trying to prove it is isomorphic to a tensor product to begin with, however, you can construct the inverse another way.

Define $h_1 : M \times N \rightarrow M \otimes (N \oplus L)$ by $h_1(m, n) = m \otimes (n, 0)$ and $h_2 : M \times L \rightarrow M \otimes (N \oplus L)$ by $h_2(m, l) = m \otimes (0, l)$, check that these are bilinear, and thus give you maps $\hat{h_1} : M \otimes N \rightarrow M \otimes (N \oplus L)$ and $\hat{h_2} : M \otimes L \rightarrow M \otimes (N \oplus L)$. Then check that $h : (M \otimes N) \oplus (M \otimes L)$ defined by $h(m_1 \otimes n, m_2 \otimes l) = \hat{h_1}(m_1 \otimes n) + \hat{h_2}(m_2 \otimes l)$ gives you an inverse.