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Mar 16, 2023 at 5:32 comment added sudgy The original motivation for this question was about utilizing other possible constructions. I've heard people say that you can directly get the tensor algebra from the general adjoint functor theorem so I was wondering how you could derive the usual properties of the tensor algebra without the usual construction.
Mar 16, 2023 at 5:27 comment added LSpice I deleted my second comment; I forgot that $T(V)$ is not a ‘bare’ algebra, but an algebra with a distinguished map $V \to T(V)$. So I do not have a technical objection to this approach. But it still seems philosophically like you are showing, without explicitly referring to tensors, that the usual construction is the only one that could work—at which point what have we gained in understanding over just carrying out the usual construction anyway, since we have to do something like that to show that this universal object exists?
Mar 16, 2023 at 5:14 comment added sudgy This is using only the universal property. It doesn't utilize any pre-existing construction at all.
S Mar 16, 2023 at 5:06 review First answers
Mar 16, 2023 at 5:25
S Mar 16, 2023 at 5:06 history edited LSpice CC BY-SA 4.0
`\oplus` -> `\bigoplus`
Mar 16, 2023 at 3:43 comment added LSpice I thought your goal was to prove the graded-ness by using only the universal property. As I mentioned, and indeed as you said in your original question, it is indeed easy to prove graded-ness if you are willing either to construct the tensor algebra, or at least to use the tensor algebra as a model target of $R$-module homomorphisms from $V$.
Mar 16, 2023 at 3:13 history edited sudgy CC BY-SA 4.0
added 58 characters in body
Mar 16, 2023 at 3:07 vote accept sudgy
S Mar 16, 2023 at 3:07 review First answers
Mar 16, 2023 at 3:41
S Mar 16, 2023 at 3:07 history answered sudgy CC BY-SA 4.0