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nLab: I guess a left module should be an enriched functor not an enriched presheaf.

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    • CommentRowNumber1.
    • CommentAuthorX.S.Gau
    • CommentTimeMay 19th 2014

    In this page http://ncatlab.org/nlab/show/module#InEnrichedCategory, They defined the left module over a monoid object A as an enriched presheaf. However, consider the modules over a ring, the left module should be an Ab−\mathbf{Ab}-enriched functor. Is it a flaw or there is some reasons.

    The same thing happens in http://ncatlab.org/nlab/show/bimodule, They defined the C−D−C-D-bimodule as a V−V-enriched functor

    C op⊗D⟶VC^{\mathrm{op}}\otimes D\longrightarrow V

    However, this is not the case in the usual use of bimoudles over two rings since a R−S−R-S-bimodule, where R,SR,S are two rings, is actually a Ab−\mathbf{Ab}-enriched functor

    R⊗S op⟶VR\otimes S^{\mathrm{op}}\longrightarrow V
    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeMay 20th 2014

    Welcome to the forum! FYI, you can make links to the nlab from here with the same syntax you would use for linking between pages on the lab, e.g. [[bimodule]] makes bimodule.

    As for your question, I think it depends on how you regard a ring as an Ab-enriched category. Is the product rsr s regarded as the composite →r→s\xrightarrow{r}\xrightarrow{s} or the composite →s→r\xrightarrow{s}\xrightarrow{r}?

    • CommentRowNumber3.
    • CommentAuthorX.S.Gau
    • CommentTimeMay 20th 2014
    • (edited May 20th 2014)

    Thanks for your information.

    I think in traditional notation, rsr s is the abbreviation of r∘sr\circ s, that is the composite →s→r\xrightarrow{s}\xrightarrow{r}. What’s more, in the paragraph just next to the definition, they say a left module is equivalently an Ab−Ab-enriched functor.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeMay 20th 2014

    In a ring , rsr s is not an abbreviation, but is the product operation. When regarding a ring RR equivalently as a pointed connected AbAb-enriched category *→BR\ast \to \mathbf{B}R there are two equivalent ways in doing so: each element r∈Rr \in R is regarded as a morphism r^\hat r in BR\mathbf{B}R, but then one may decide either way if composition of morphisms r^s^=r^∘s^\hat r \hat s = \hat r \circ \hat s corresponds to the product rsr s or to the product srs r. Both conventions have their justification, but in either case they are (just) conventions.

    All that non-withstanding, I agree that the entry could use at least a remark on this issue, if this doesn’t become clear. Somebody hopefully is so kind to add it.

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeMay 20th 2014

    3: yes, for the ordinary composition, a left module is simply Ab-functor and a right module is simply Ab-presheaf. Then CC-DD-bimodule is C op⊗DC^{op}\otimes D-presheaf. Cf.

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