Timeline for Primitive recursive arithmetic via universal algebra

9 events

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Nov 7 at 12:21	comment	added	Nulhomologous		Can one explain how to prove the assertion that the morphisms $1\to N$ correspond bijectively to standard natural numbers? I understand one can assign, using the Lawvere theory morphism $f$, a unique natural number to any such morphism, and that one can construct a morphism for any natural number (inductively, as $s\circ \dots \circ s\circ 0$). But I am not sure how one can prove that the functorial map $\hom_T(1, N) \to \hom(1, \mathbb{N})\cong \mathbb{N}$ is injective, so a bijection.
Apr 24, 2020 at 9:15	history	edited	Todd Trimble	CC BY-SA 4.0	fixed broken link
Apr 24, 2020 at 9:08	history	edited	Todd Trimble	CC BY-SA 4.0	updated to include precise references
Dec 8, 2016 at 8:33	comment	added	goblin GONE		Actually, that's exactly how I think about it; see example 2 here.
Dec 6, 2016 at 12:48	comment	added	Todd Trimble		@goblin May be just the way I think. For an NNO, I often write $1 \stackrel{0}{\to} \mathbb{N} \stackrel{s}{\leftarrow} \mathbb{N}$ to suggest we are thinking of the $\mathbb{N}$ in the middle as equipped with coproduct inclusions, realizing an isomorphism $1 + \mathbb{N} \cong \mathbb{N}$.
Dec 6, 2016 at 11:56	comment	added	goblin GONE		Todd, is there any significance to $s \times 1_A$ and $g$ being backwards?
Mar 31, 2015 at 4:22	vote	accept	Rex Butler
Feb 21, 2015 at 23:47	vote	accept	Rex Butler
Feb 24, 2015 at 21:48
Feb 20, 2015 at 5:14	history	answered	Todd Trimble	CC BY-SA 3.0