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Let $(F, \phi): \mathscr{A}\to \mathscr{B}$ a lax-functor between 2-categories. In the setting of 2-categories the axiom of lax-functor become:

Ul) $1: F(f)=F(f)\circ 1_{F(X)}\xrightarrow{1\circ \phi_X}F(f)\circ F(1_X)\xrightarrow{\phi_{f, 1}} F(f\circ 1_X)= F(f)$.

For $f: X \to Y$ in $\mathscr{A}$.

Ur) $1: F(f)= 1_{F(Y)}\circ F(f)\xrightarrow{\phi_X\circ 1} F(1_Y)\circ F(f)\xrightarrow{\phi_{1, f}} F( 1_Y \circ f)= F(f)$.

For $f: X \to Y$ in $\mathscr{A}$.

UA) The compositions $F(h)\circ F(g)\circ F(f) \xrightarrow{\phi_{h, g}\circ 1} F(h\circ g)\circ F(f) \xrightarrow{\phi_{h\circ g, f}\circ 1} F(h\circ g\circ f)$ and

$F(h)\circ F(g)\circ F(f) \xrightarrow{1\circ \phi_{g, f}} F(h)\circ F(g\circ f) \xrightarrow{\phi_{h, g\circ f}} F(h\circ g\circ f)$ are equal.

For componibile morphisms $h, g, f$.

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I call $(F, \phi)$ normal (or unitary) if in the axioms $(Ul)$ and $(Ur)$ above all arrows are identities.
Gived $(F, \phi)$ (general) I define a normal lax.functor $(\tilde{F}, \tilde{\phi})$ that is the some of $F$ on objects and on non-identity morphisms, with of course $\tilde{F}(1_X)=1_{F(X)}$, $\widetilde{\phi}_{X}= 1: \tilde{F}(1_X)\to 1_{F(X)} $, and with $\tilde{\phi}_{g, f}$ defined as:

$ \phi_{g, f}: F(g)\circ F(f)\to F(g\circ f)$ if $f$ and $g$ arent identity, and the obvious identity if $f$ or $g$ is a identity. I checked (easly) that axiom $(UA)$ is true for $(\tilde{F}, \tilde{\phi})$, then $(\tilde{F}, \tilde{\phi})$ is a lax.funtor.

I ask if this (very easy normalization) is just know in literature ( I dont know), and if is right (I'm sure its right, but I'm no too sure of myself).

Edit: If in $(Ul)$ all arrows are identies, this imply the some in $(Ur)$ (i.e. $(F, \phi)$ is normal)?

My motivations is a generalization of the J.W. Gray concept of quasi-functors to lax.functors (Gray gived this definition for 2-functors).

EDit As Jonathan Chiche observed, these is the obvious condition than a lax.functor induce hom-functors between hom-categories (I am ashamed for this rough oversight). Anyway if the canonical morphism $\phi_X: 1_{FX}\Rightarrow F(1_X)$ is a isomorphism then the mine definition preserving this funtorialiy, only need a easy correction based on the following observation: the cells of type $\sigma: 1_{FX}\Rightarrow W$ are in bijections with the cells of type $\sigma': F(1_X)\Rightarrow W$ and the cells like $\tau: W\Rightarrow 1_{FX}$ are in bijections with the cells of type $\tau': W\Rightarrow F(1_X)$. With this correction I seems that also the naturality of $\tilde{\phi}$ work well.

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It seems to me that your question woud be clearer if you stated more precisely the axioms for lax functors. Anyway, while I am unsure as to what you have in mind, I guess that "lax functors" and "normal lax functors" have the same meaning for you and me. You may therefore be interested in Lemma 4.2. and the beginning of the proof of Theorem 6.3. of the paper Nerves and classifying spaces for bicategories by Carrasco, Cegarra and Garzón. Hope this helps.

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  • $\begingroup$ Of course the axioms are the classical: J.Benabou, Introduction to bicategories, (M1) and (M2) p. 30. Thank your for the reference. A lax functors (for me) is what J.Benabou call morphisms (of bicategories) or what J.W. Gray call "pseudo.functors" in its "Formal category theory" LNM 391, 1974. A lax.functor $F$ (for me) is normal if the canonical cell $1_{F(X)}\Rightarrow F(1_X)$ is the identity. $\endgroup$
    – Buschi Sergio
    Commented Nov 1, 2012 at 10:55
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    $\begingroup$ Well, I know what the axioms are, but part of my point is that some of these axioms are missing in your question. For instance, you don't mention the fact that lax functors induce functors between the categories of 1-cells. I don't think this condition is automatically satisfied as soon as the ones you mention are, even in the realm of (strict) 2-categories. $\endgroup$
    – Jonathan Chiche
    Commented Nov 1, 2012 at 11:50
  • $\begingroup$ THank you very much Jonatan (I'm no a researcher, I study alone after job, and sometime I'm tired and I take misunderstandings for stress or enthusiasm). Anyway for pseudonfunctors I seems that is possible correct the HOm-functors (functors inducted between Hom-categories) $\endgroup$
    – Buschi Sergio
    Commented Nov 1, 2012 at 17:37
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I think Jonathan is right: your "normalized" $\tilde{F}$ may not be a lax functor, because you may not be able to extend it functorially on the 2-cells. Suppose that $\mathcal{A}$ contains a 1-cell $f:X\to X$ with a 2-cell $\alpha: f\to 1_X$, whereas $\mathcal{B}$ contains no 2-cells from $F(f)$ to $1_{F(X)}$.

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  • $\begingroup$ THis is no possible, consider $\phi_X^{-1}\ast F(\alpha): F(f) \to F(1_X)\to 1_{F(X)}$, in the last Edit on my inital question I admit that the "Lax" condiction is too strong, then I limitate the conjecture on pseudo-functors (its enought pseudo only about unity condiction). LEt me know if languaage is not understable. $\endgroup$
    – Buschi Sergio
    Commented Nov 6, 2012 at 16:12
  • $\begingroup$ For a pseudofunctor this normalization is right (by the a well know coherence theorem), I used this in a my old answere to a your question abot Gray tensor product (i'm still studing about..). But I think that condiction is right also for lax functors pseudo on unity $\phi_X: 1_{F(AX)}\cong F(1_X)$. $\endgroup$
    – Buschi Sergio
    Commented Nov 6, 2012 at 16:17
  • $\begingroup$ Yes, for a normal lax functor it may work. $\endgroup$
    – Mike Shulman
    Commented Nov 7, 2012 at 21:09

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