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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.

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    $\begingroup$ The inverse is given by $f \mapsto (f\circ j_i)$, where $j_i \colon V_i \to V_1 \oplus \dotsb \oplus V_n$ is the canonical injection. $\endgroup$ Commented Sep 18, 2013 at 14:58

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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.

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  • $\begingroup$ Just to clarify the above is a rigorous way of saying that we go to each $f_i(v_i)$ and send it to $V_1\bigoplus \cdots \bigoplus V_n$ we then add the results of this giving us $(v_1, \cdots , v_n)$ and then we just have to compose our original operation with $f$ to get $f(v_1, \cdots , v_n)$ $\endgroup$
    – Ben
    Commented Sep 18, 2013 at 15:21
  • $\begingroup$ @Ben, correct . $\endgroup$
    – Norbert
    Commented Sep 18, 2013 at 21:15

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