- #1
NihilTico
- 32
- 2
A couple of notes first:
1.
[itex]\hom_{A}(-,N)[/itex] is the left-exact functor I'm referring to; Lang gives an exercise in the section preceeding to show this.
2.
This might be my own idiosyncrasy but I write TFDC to mean 'The following diagram commutes'
3.
Titles are short, so I know that the hom-functor here isn't actually taking a coproduct to a product in the same category. The coproduct [itex]\bigoplus M_i[/itex] lies in the category [itex]\text{Mod}(A)[/itex] while the product [itex]\prod\hom_{A}\left(M_i,N\right)[/itex] lies in [itex]\bf{Ab}[/itex].
I don't think I saw this before, but on page 131 of Lang's Algebra (3rd edition) he writes (NOTE: I'm dropping the indexing set [itex]I[/itex] unless I feel it is necessary for clarity) what generalizes naturally to [itex]\hom_{A}\left(\bigoplus M_{i},N\right)\approx\prod\hom_{A}\left(M_i,N\right)[/itex]. Where the [itex]M_i[/itex] and [itex]N[/itex] are $A$-modules. But he goes about demonstrating it for the case of two [itex]A[/itex]-modules, curiously.
To be a thorough as possible, is it the case, since the coproduct is the initial object in the category of tuples [itex]\left(C,\left\{f_{i}\colon{M_i}\to{C}\right\}_{i\in{I}}\right)[/itex] (where the [itex]f_i[/itex] are [itex]A[/itex]-homomorphisms) meaning that a morphism [itex]f\colon\left(\bigoplus M_i,\left\{\jmath_{i}\colon{M_i}\to{\bigoplus {M_i}}\right\}_{i\in{I}}\right)\to\left(C,\left\{f_{i}\colon{M_i}\to{C}\right\}_{i\in{I}}\right)[/itex] is a unique [itex]A[/itex] homomorphism [itex]h\colon \bigoplus M_i\to{C}[/itex] such that [itex]\forall i\in{I}[/itex] TFDC:
[itex] \begin{array}{cccc} & \jmath_{i} & \bigoplus M_{i}\\ M_{i} & \nearrow & \downarrow\\ & \searrow & \ \downarrow h\\ & f_{i} & C \end{array}[/itex]
that this is totally obvious, since [itex]h[/itex] is uniquely determined by the family [itex]\left\{f_i\right\}_{i\in{I}}[/itex]? In particular, the problem becomes—more or less—one of the existence of a group homomorphsim [itex]q\colon\hom_{A}\left(\bigoplus M_{i},N\right)\to\prod\hom_{A}\left(M_i,N\right)[/itex].
[/B]
Not applicable.
[/B]
Not so much as an attempt, as an observation.
Clearly, any [itex]h\colon\bigoplus M_i\to{N}[/itex] induces a unique family of [itex]A[/itex]-homomorphisms given by [itex]\left\{h_i\colon M_i\to{N}\right\}_{i\in{I}}[/itex] ; after all, if another family had this [itex]h[/itex] as well, then the families would be equal as well by definition of the coprouct. Similarly, for any family in the product on the right, there is a unique [itex]h[/itex] from the coproduct to [itex]N[/itex]. Isomorphisms in [itex]\bf{Ab}[/itex] are bijective so this is enough by defining [itex]q(h)=\left(h_i\right)_{i\in{I}}[/itex].
Right?
If the above is correct, does it speak to something deeper about hom-functors that I'd be better served finding in Mac Lane's CWM?Thanks
1.
[itex]\hom_{A}(-,N)[/itex] is the left-exact functor I'm referring to; Lang gives an exercise in the section preceeding to show this.
2.
This might be my own idiosyncrasy but I write TFDC to mean 'The following diagram commutes'
3.
Titles are short, so I know that the hom-functor here isn't actually taking a coproduct to a product in the same category. The coproduct [itex]\bigoplus M_i[/itex] lies in the category [itex]\text{Mod}(A)[/itex] while the product [itex]\prod\hom_{A}\left(M_i,N\right)[/itex] lies in [itex]\bf{Ab}[/itex].
Homework Statement
I don't think I saw this before, but on page 131 of Lang's Algebra (3rd edition) he writes (NOTE: I'm dropping the indexing set [itex]I[/itex] unless I feel it is necessary for clarity) what generalizes naturally to [itex]\hom_{A}\left(\bigoplus M_{i},N\right)\approx\prod\hom_{A}\left(M_i,N\right)[/itex]. Where the [itex]M_i[/itex] and [itex]N[/itex] are $A$-modules. But he goes about demonstrating it for the case of two [itex]A[/itex]-modules, curiously.
To be a thorough as possible, is it the case, since the coproduct is the initial object in the category of tuples [itex]\left(C,\left\{f_{i}\colon{M_i}\to{C}\right\}_{i\in{I}}\right)[/itex] (where the [itex]f_i[/itex] are [itex]A[/itex]-homomorphisms) meaning that a morphism [itex]f\colon\left(\bigoplus M_i,\left\{\jmath_{i}\colon{M_i}\to{\bigoplus {M_i}}\right\}_{i\in{I}}\right)\to\left(C,\left\{f_{i}\colon{M_i}\to{C}\right\}_{i\in{I}}\right)[/itex] is a unique [itex]A[/itex] homomorphism [itex]h\colon \bigoplus M_i\to{C}[/itex] such that [itex]\forall i\in{I}[/itex] TFDC:
[itex] \begin{array}{cccc} & \jmath_{i} & \bigoplus M_{i}\\ M_{i} & \nearrow & \downarrow\\ & \searrow & \ \downarrow h\\ & f_{i} & C \end{array}[/itex]
that this is totally obvious, since [itex]h[/itex] is uniquely determined by the family [itex]\left\{f_i\right\}_{i\in{I}}[/itex]? In particular, the problem becomes—more or less—one of the existence of a group homomorphsim [itex]q\colon\hom_{A}\left(\bigoplus M_{i},N\right)\to\prod\hom_{A}\left(M_i,N\right)[/itex].
Homework Equations
[/B]
Not applicable.
The Attempt at a Solution
[/B]
Not so much as an attempt, as an observation.
Clearly, any [itex]h\colon\bigoplus M_i\to{N}[/itex] induces a unique family of [itex]A[/itex]-homomorphisms given by [itex]\left\{h_i\colon M_i\to{N}\right\}_{i\in{I}}[/itex] ; after all, if another family had this [itex]h[/itex] as well, then the families would be equal as well by definition of the coprouct. Similarly, for any family in the product on the right, there is a unique [itex]h[/itex] from the coproduct to [itex]N[/itex]. Isomorphisms in [itex]\bf{Ab}[/itex] are bijective so this is enough by defining [itex]q(h)=\left(h_i\right)_{i\in{I}}[/itex].
Right?
If the above is correct, does it speak to something deeper about hom-functors that I'd be better served finding in Mac Lane's CWM?Thanks