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Given the (external) direct sum of $R$-modules: $V= \bigoplus V_i, W= \bigoplus W_j$, will $\hom(V,W) \cong \bigoplus_{i,j} \hom(V_i,W_j)?$

I know that $\hom(V,\bigoplus_j W_j) \cong \bigoplus_j \hom(V,W_j)$, the isomomorphism given by $\Phi(f) \mapsto (f\circ \pi_1,\cdots, f\circ \pi_s)$. Would then the same argument give us $\hom(V,W_j) \cong \bigoplus_i \hom(V_i,W_j) $? I'm not sure about this, since a $f\in \hom(V,W_j)$ is given by $f(v_1,\cdots, v_k) = w_j \in W_j.$

How to proceed with this? Is it actually true?

user2345678
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  • It's true if both sums are finite. If both are infinite, you get a double direct product. Recall that for a finite number of modules direct sum and direct product are "the same." The relevant results should be in any good graduate algebra text. – Chris Leary Oct 03 '19 at 19:16
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    It's $\prod_i\bigoplus_j\text{Hom}(V_i,W_j)$. – Angina Seng Oct 03 '19 at 19:36

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In the case where both sums are finite, then you are exactly right (As Chris points out). However, as Lord Shark points out, if either of the sums are infinite things can get weird. In general:

$$\text{Hom} \left ( \bigoplus_i V_i, \prod_j W_j \right ) \cong \prod_i \prod_j \text{Hom}(V_i, W_j)$$

However, since direct sums and direct products coincide when the index set is finite, your answer is correct in that case.

As for why...

Say $f : \bigoplus_i V_i \to \prod_j W_j$.

Then $f \circ \iota_i : V_i \to \prod_j W_j$ (where $\iota_i : V_i \to \bigoplus_i V_i$)

Then $\pi_j \circ f \circ \iota_i : V_i \to W_j$ (where $\pi_j : \prod_j W_j \to W_j$).

Thus, pairing these up, $(\pi_j \circ f \circ \iota_i)_{i, j} : \prod_i \prod_j \text{Hom}(V_i, W_j)$.

Since this construction is uniform in $f$, it defines one direction of the desired isomorphism. I will leave it to you to prove the other direction.

I hope this helps ^_^

HallaSurvivor
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