In synthetic differential geometry, an object $M$ verifies the *Wraith axiom* if for all functions $\tau:D\times D\to M$ which are constant on the axes $\tau(d,0)=\tau(0,d)=\tau(0,0)$ for all $d\in D$, there's a unique factorization through the multiplication map, i.e there's a unique function $t:D\to M$ such that $t(d\_1\cdot d\_2)=\tau(d\_1,d\_2)$.


What is the geometric/physical intuition behind this axiom? What is the analogue in the category of smooth manifolds?




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**Update.** Following the answers I think I should add some motivation. Lavendhomme's book defines the commutator of vector fields as $\tau(d\_1,d\_2)=Y\_{-d\_2}X\_{-d\_1}Y\_{d\_2}X\_{d\_1}$. Because this is constant on axes, it factors through the multiplication map to give a vector field $t$ charaterized by $t(d\_1d\_2)=\tau(d\_1,d\_2)$. I understand $t$ is desirable since it is a vector field, but I don't know how to geometrically interpret its characterizing property. For instance, why not consider the vector field $\tau(d\_1,d\_1)$ given by precomposing the diagonal? This motivated my question.



Intuition and analogue of Wraith axiom from synthetic differential geometry
Axiom W is about the behaviour of the second tangent bundle - it ensures that the vertical bundle of the tangent bundle, $V(M) \subseteq T\circ T(M)$, where $V(M) = T(p)^{-1}(0)$, decomposes as the pullback of the projection $p\_M: T(M) \to M$ along itself. The map $[\bullet, M]:[D,M] \to [D \times D, M]$ would be written fiberwise as $\ell(v) = \frac{d}{dt}\_{t = 0} (vt)$. If you look at [Robin Cockett and Geoff Cruttwell's](https://www.mta.ca/uploadedFiles/Community/Bios/Geoff_Cruttwell/sman3.pdf) first paper on tangent categories, you can see they spend a fair bit of time talking about the universality of the vertical lift and its relationship with the Lie bracket.


You may also want to look at the definition of the core of a [double vector bundle](https://arxiv.org/pdf/1705.01017.pdf) - axiom W can also be read as saying the core of the second tangent bundle of M is $TM$. 


Edit: If you look at the answer [here](https://math.stackexchange.com/questions/1687095/why-the-vertical-lift-of-a-vector-bundle-takes-values-on-the-vertical-bundle), you can see how the lift from the fibred product $TM \times\_M TM \to T^2(M)$ is written. Using infinitesimals, you would write $\gamma,\beta:D\to M, \gamma(0)=\beta(0)$ is sent to the map $D\times D \to M$ which is given by $d\_1,d\_2 \mapsto \gamma(d\_1) + \beta(d\_1d\_2)$. The condition that this be the kernel of $T(p)$ is provable from property W (and a good exercise).


The square root business you mentioned looks to be how you would show property $W$ holds in the category of smooth manifolds (when rewritten as the universality of the vertical lift from the tangent category axioms).


