I'm confused with what this bijection is and how I would find it. So:
Given vector spaces $V_1,V_2,\cdots,V_n,W$ and linear mappings $f_i : V_i \to W$ then we can form a new linear mapping $f:V_1\bigoplus \cdots \bigoplus V_n \to W$ by $f(v_1,v_2, \cdots , v_n) = f_1 (v_1) + f_2(v_2) + \cdots +f_n(v_n)$. In this way we even get a bijection:
$$\text{Hom}(V_1,W) \times \cdots \times \text{Hom}(V_n,W) \xrightarrow{\sim} \text{Hom}(V_1\bigoplus \cdots \bigoplus V_n,W) $$
I understand why we can get another linear map by adding them all together but how can i show that there is a bijection between the two sets above? I'm not even convinced there is so I think I am missing something basic.
I've tried saying the bijection is just $(f_1, f_2, \cdots ,f_n) \to f$ with $f$ as above but have been unable to show this is surjective or injective.
Best Answer
Define maps $$ I:\prod\limits_{i=1}^n\hom(V_i,W)\to\hom\left(\bigoplus\limits_{i=1}^n V_i,W\right),\prod\limits_{i=1}^n f_i\mapsto\left(\bigoplus\limits_{i=1}^n v_i\mapsto\sum\limits_{i=1}^n f_i(v_i)\right) $$ $$ J:\hom\left(\bigoplus\limits_{i=1}^n V_i,W\right)\to\prod\limits_{i=1}^n\hom(V_i,W): f\mapsto\prod\limits_{i=1}^n (f\circ j_i) $$ where $j_i:V_i\to\bigoplus\limits_{i=1}^n V_i$ is a natural embedding. It is straightforward to check that $I$ and $J$ are inverse to each other.