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Let $\mathcal{V}$ be a semi-monoidal category, meaning it satisfies the axioms of a monoidal category except missing a unit and the unit axiom. One could then still go about defining a $\mathcal{V}$-category by dropping the requirement of having unit morphisms.

One concern is that without unit morphisms there is no way to define an underlying category $\mathcal{C}_0$ associated to a $\mathcal{V}$-category $\mathcal{C}$. However other than that it seems to me that other parts of the theory makes sense.

Has such categories been considered previously anywhere in literature? I can find several mentions of semi-monoidal category e.g. https://arxiv.org/abs/math/0507349 but no mentions of semi-enriched categories. Of particular interest to me would be if there is a version of the enriched Yoneda lemma for such categories.

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    $\begingroup$ Categories without units are called semicategories. You can enrich a semicategory in a semigroupal category, which is what you describe. The Yoneda lemma is subtle with semicategories, but see On regular presheaves and regular semi-categories. $\endgroup$
    – varkor
    Commented Feb 4, 2022 at 20:48
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    $\begingroup$ I'm happy accepting this as an answer as it addresses everything in my question. $\endgroup$
    – Bjorn
    Commented Feb 5, 2022 at 22:27

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Expanding upon my comment: categories without units are called semicategories. You define a notion of semicategory enriched in a semigroupal category, which is what you describe. The Yoneda lemma is subtle with semicategories, but see On regular presheaves and regular semi-categories. However, note that the authors work with semicategories enriched in a monoidal category: this is because, despite the definition of enriched semicategory and semifunctor not needing a unit in $\mathcal V$, a unit is necessary to define an enriched notion of natural transformation between semifunctors. I am not aware of a reference that explicitly develops the theory of semicategories enriched in semigroupal categories.

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  • $\begingroup$ what do you mean by enriching a category in a semimonoidal category? my understanding of enriching a cat in a monoidal cat is to give an enriched cat on the same objects whose 'underlying' cat is the original one. but as pointed out in the question, there is no notion of 'underlying cat' of a cat enriched in a semimonoidal cat, and in absence of this i don't see what it should mean to be an enrichement. $\endgroup$
    – Jonas Frey
    Commented Sep 4, 2023 at 22:47
  • $\begingroup$ @JonasFrey: there are multiple ways to view the concept of enriched category. One is as structure on a category, in which case one expects to be able to recover the underlying category from the enriched category. However, another perspective is simply a category-like structure whose homs form not a set, but the object of some other category. Here, a priori, there is no reason to expect this notion of enriched category to have an underlying category, though it turns out to be true by homming out of the unit. (1/3) $\endgroup$
    – varkor
    Commented Sep 5, 2023 at 6:20
  • $\begingroup$ Enrichment in semigroupal/semimonoidal categories is motivated by the second perspective and, here, it turns out that there is no notion of underlying semicategory. So the first perspective on enrichment does not generalise to enrichment in semimonoidal categories. However, one can find examples of such enriched semicategories (as in the linked paper), which suggests the definition is reasonable. (2/3) $\endgroup$
    – varkor
    Commented Sep 5, 2023 at 6:22
  • $\begingroup$ However, I think the term "enrichment" suggests the first perspective, so I think it would be fair to say there is no concept of "enrichment of a semicategory over a semimonoidal category". There is just the concept of "enriched semicategory". Most likely you are pointing out the existing wording in my answer is misleading in this respect? I will tweak the wording accordingly, thank you. (3/3) $\endgroup$
    – varkor
    Commented Sep 5, 2023 at 6:24
  • $\begingroup$ yes, my question was specifically aimed at the previous wording, and I was wondering if I was missing something. good to see that we're on the same page! $\endgroup$
    – Jonas Frey
    Commented Sep 5, 2023 at 17:07

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