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created immersion of smooth manifolds .
Stated also the definition that $f : X \to Y$ is an immersion precisely if the canonical morphism
$T X \to X \times_Y T Y =: f^* Y$is an injection.
This style of writing the conditon I have now also added to submersion (where this canonical morphism is a surjection) and to local diffeomorphism (where it is an iso).
This way of stating the condition makes most manifest that with respect to the infinitesimal cohesion $i :$ Smooth∞Grpd $\hookrightarrow$ SynthDiff∞Grpd we have
immersion $\Leftrightarrow$ formally unramified morphism
submersion $\Leftrightarrow$ formally smooth morphism
local diffeomorphism $\Leftrightarrow$ formally etale morphism
(Note that \eqqcolon
produces $\eqqcolon$)
Thanks!
I sort of knew this, but I always forget and am to lazy to look it up. I will try to remember it from now on.
I need to add more discussion to this:
formally unramified in the smooth context implies immersion, but superficially is stronger: it says not just that the map on tangent spaces is an injection, but also that the map on all “higher order tangents” (formal curves, as it were) is an injection. That shouldn’t change the statement, but needs to be discussed.
$X \times_Y T Y =: f^* Y$
Firstly I had some reservations against this notation (since forming ” $f^* f$ ” or the like won’t lead to any good) - but it has something to it. Has this object a name in the literature?
How are higher order tangents $\tau$ defined in the smooth context? If they were just $\tau=Hom(D,-)$ for the infinitesimal smooth loci (and not just the infinitesimal standard interval) then I think formally unramified coincides with immersion if every such $D$ is atomic (in that $\tau$ is a right adjoint).
Note that on the page in question it reads $X \times_Y T Y \eqqcolon f^* T Y$ and this is the standard notation for the pullback of a bundle by a smooth map. Indeed, it’s the pullback of the lower-right corner of the diagram just below.
Ok, I just read the thread here.
How are higher order tangents defined in the smooth context?
Good that you ask. I should have told you about that earlier.
Here is how it goes:
we say a (smooth) infinitesimal thickened point is an object in the opposite category of the full sub-category of $\mathbb{R}$-algebras
$InfinitesimalPoint \hookrightarrow Alg_{\mathbb{R}}^{op}$on those whose uderlying vector space is of the form $\mathbb{R} \oplus V$ with $V$ a finite dimensional $\mathbb{R}$-vector space and such that all elements $\epsilon \in V$ are niltpotent in the given algebra structure, i.e. such that there is $n \in \mathbb{N}$ with $\epsilon^n = 0$.
The basic example is the $\mathbb{R}$-algebra version of the ring of dual numbers $\mathbb{R} \oplus \epsilon \mathbb{R}$.
By the Hadamard lemma this can be thought of as $C^\infty(\mathbb{R})/\langle x^2\rangle$, where the quotient is by the ideal of smooth functions that a products with the function $x \mapsto x^2$. This way one sees that all these infinitesimally thickened points can uniquely be thought of as smooth loci, formal duals of smooth algebras (there is a unique smooth algebra structure on an $\mathbb{R}$-algebra of the above form).
So write $\mathbb{D}$ for the infinitesimal point corresponding to the ring of dual numbers. By definition we say that the ring of dual numbers is the smooth algebra of smooth functions on $\mathbb{D}$:
$C^\infty(\mathbb{D}) := C^\infty(\mathbb{R})/\langle x^2\rangle \,.$This means that a morphism of smooth loci $v : \mathbb{D} \to X$ is equivalently (by definition) a morphism of smooth algebras
$C^\infty(\mathbb{D}) \leftarrow C^\infty(X) : v^* \,.$Compare this to how a point $x : * \to X$ is indeed the same thing as an algebra homomorphism
$\mathbb{R} = C^\infty(*) \leftarrow C^\infty(X) : x^*$(given by evaluation of functions on that point).
Exercise: convince yourself, for $X$ a smooth manifold, that morphisms $v : \mathbb{D} \to X$ are naturally identified with pairs: a point $x \in X$ together with a choice of tangent vector $v \in T_x X$.
(Hint: you can just work with bare $\mathbb{R}$-algebras for this, don’t need to regard them as smooth algebras.)
So this says that $T X = Hom(\mathbb{D}, X)$ as sets, and also as smooth spaces if we form the internal hom $T X = [\mathbb{D}, X]$.
Now let $W$ be any other “algebra of functional on an infinitesimally thickened point” as above (some algebra of the form $\mathbb{R} \oplus V$ with $V$ nilpotent). Write $D$ for the corresponding infinitesimally thickened point. Then a “higher order tangent” in $X$ as given by this $D$ is a morphism of smooth loci
$\nu : D \to X \,,$hence an algebra homomorphism
$W \leftarrow C^\infty(X) : \nu^* \,.$Once you are happy with the above exercise, it is immediate to see in which sense and way these morphisms are “higher order tangents” on $X$.
I have to quit now and talk to people in Göttingen. I’ll try to get back to you later this eveneing.
Thanks, Urs for showing this to me - and for introducing me to the subject of synthetic differential geometry in general!
Re #1, how does the concept of ’formal immersion’ fit in? This is
A formal immersion $F$ from $M$ to $N$ is an injective bundle map $T M \to T N$.
It seems to be easier to handle than the plain immersions, though in some case collections of each are weakly homotopy equivalent.
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