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Stream: learning: questions

Topic: Finite Biproducts Without Pointedness

Reed Mullanix (Feb 26 2021 at 20:27):

I've been reading through the (excellent) paper "Biproducts without Pointedness"[1], and was trying to generalize the ideas there to finite biproducts, rather than binary biproducts. The definition is easy enough to extend, however, I keep getting caught up when trying to show that binary biproducts + a zero object implies finite biproducts. Does anyone know of any references that explore this?

[1] https://arxiv.org/abs/1801.06488

Reed Mullanix (Feb 26 2021 at 20:36):

Of course, the _second_ I posted this I thought of a potential solution :upside_down:. The definition presented by Karvonen lists has the following condition: $i_1 \circ \pi_1 \circ i_2 \circ \pi_2 = i_2 \circ \pi_2 \circ i_1 \circ \pi_1$. I think by strengthening this to $i_1 \circ f \circ \pi_1 \circ i_2 \circ g \circ \pi_2 = i_2 \circ g \circ \pi_2 \circ i_1 \circ f \circ \pi_1$ we can show that binary biproducts + zero objects imply finite biproducts. Does that sound horribly wrong to anyone?

JS PL (he/him) (Feb 26 2021 at 20:38):

@Martti Karvonen might be the person to ask! I think he's usually very active here.

Reed Mullanix (Feb 26 2021 at 20:39):

I didn't even think to check if he was around, that's fantastic! :smile:

Martti Karvonen (Feb 26 2021 at 21:02):

So, once you have a zero object the modest generalization I proposed coincides with the usual one, so we might as well work with the definition that a biproduct is a cone and a cocone that is simultaneously a coproduct, a product and $\pi_j i_k=\delta_{jk}$ (i.e. identity if $j=k$ and zero otherwise). Then to see that binary biproducts + zero objects induce n-ary biproducts I'd argue as follows: define $\bigoplus_{i=1}^n A_i$ for instance by $(A_1\oplus A_2).....)\oplus A_n$ (i.e. add everything together pairwise with the leftmost bracketing). The usual proofs that binary products & terminal object->finite products (and its dual) imply that the end result is simultaneously a product and a coproduct when equipped with the obvious projections and injections to A_i. Moreover, the building blocks of this sum satisfying the desired equations imply that $\pi_j i_k=\delta_{jk}$ still holds so we are done.

Martti Karvonen (Feb 26 2021 at 21:09):

I think I now see where your question is coming from: if you prefer a proof that goes more directly to the definition I proposed, check Lemma 3.2. of the paper: injecting from a different half of a biproduct and then projecting to the other is always a zero morphism, so when you iterate pairwise sums, this guarantees that all the idempotents on the end-result that come from a projection and an injection commute pairwise, as doing distinct ones results in the zero endomorphism either way. Thus there is no need to strengthen the equation saying that $i_k\pi_k$ commutes with $i_j\pi_j$.

Reed Mullanix (Feb 26 2021 at 21:33):

Ah, that makes a lot of sense. I'll play around with things a bit more, thanks for your time!