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I am new to category theory and keep running across it while studying algebra. Whenever I see the $\mathrm{Hom}$ functor mentioned (in the context of modules), two of its basic properties are listed:

  1. $\mathrm{Hom}(\bigoplus_i{A_i},B) \cong \prod_i{\mathrm{Hom}(A_i,B)}$, and
  2. $\mathrm{Hom}(A,\prod_i{B_i}) \cong \prod_i{\mathrm{Hom}(A,B_i)}$.

Could anyone provide an intuitive explanation for why this is true? I have a hard time remembering this fact (where the product and sums ought to go on the left hand side.) Is there a concrete example which makes this clear?

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  • $\begingroup$ Shouldn't it be Hom$(\bigoplus A_i,B)\cong\prod$Hom$(A_i,B)$. $\endgroup$ – Stefan Hamcke Jul 25 '13 at 1:10
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    $\begingroup$ The $\prod$ and $\bigoplus$ on the left sides of your isomorphisms should be interchanged. The $\prod$'s on the right side are OK. Once the formulas are corrected, they amount to the category-theoretic definition of coproducts ($\bigoplus$) and products ($\prod$). $\endgroup$ – Andreas Blass Jul 25 '13 at 1:25
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    $\begingroup$ I remember these laws thanks to their logical counterparts : ((A1 \/ A2) => C) <=> (A1 => C) /\ (A2 => C) and (A => (B1 /\ B2)) <=> (A => B1) /\ (A => B2) $\endgroup$ – Romuald Jul 25 '13 at 8:27
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Assume we are in the category of abelian groups (for intuition's sake).

a. $\mbox{Hom}(\bigoplus A_i,B)\simeq\prod\mbox{Hom}(A_i,B)$.

Let $\phi_i:A_i\hookrightarrow\bigoplus A_i$ be the natural inclusion (we put 0's in all the other coordinates), and if $f\in\mbox{Hom}(\bigoplus A_i,B)$, let $f_i$ be its restriction to $\phi_i(A_i)$ (Edit: If $p_i:\bigoplus A_i\to A_i$ is the natural projection, then by "restriction" I really mean $f_i=f\circ\phi_i\circ p_i$). We see then that $f=\sum f_i$. This sum is infinite, but since an element of $\bigoplus A_i$ has finite coordinates, we can see it as a well-defined function. The isomorphism above is easy then: send $f$ to $(f_i)_{i}$ (the $f_i$ are functions on $\phi_i(A_i)\simeq A_i$).

b. $\mbox{Hom}(A,\prod B_i)\simeq\prod\mbox{Hom}(A,B_i)$.

Let $\pi_i:\prod B_i\to B_i$ be the natural projection. If $f\in\mbox{Hom}(A,\prod B_i)$, then send $f$ to $(\pi_i\circ f)$; this gives an isomorphism.

Moral of the story:

A morphism in the first situation is determined by how it acts on each $A_i$, and a morphism in the second situation is determined by its image in each $B_i$.

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As Andreas Blass commented, once you correct your statements, they are essentially the definition of coproduct and product. The way I think about these is that products have projection maps, coproducts have inclusion maps. (And to remember which is which, I think of Set, where products are cartesian products and coproducts are disjoint unions. Then "projection map" and "inclusion map" are very familiar concepts.)

Then what you are saying is, "to map from all these $A_i$'s to $B$, I could just as easily embed the $A_i$'s in their disjoint union then do one big map to $B$" and "to map from A to all these $B_i$'s, I could just as easily map from $A$ into the Cartesian product of all the $B_i$'s, then project out into the individuals".

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Direct sums are colimits, and direct products are limits. The identities you mention are actually true for all limits and colimits, not just the special case of products and coproducts.

If you remember that much, it becomes a much simpler exercise to recall which way the identities go: just work out how they should look for binary Cartesian products and binary disjoint unions in Set.

Another way to recall which is which is to remember that $\text{Hom}$ is covariant in the second argument and contravariant in the first.

If you remember that you get a limit either way, then you can work out which identity is right by noting that if you apply $\text{Hom}(A, -)$ to a limit cone, you get a limit cone, and if you apply $\text{Hom}(-,B)$ to a colimit cone, you also get a limit cone.

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