I'm interested in the geometric motivation behind the bundle (or in more general framework the sheaf) of relative differentials $\Omega_{X/Y}$
of a morphism $f: X \to Y$ smooth $k$-varieties. Differential geometry provides following picture of relative differentials:
Let $f: X \to Y$ a equidimensional surjective map between connected manifolds $Y, X$
and moreover assume that every fiber $F:= f^{-1}(y) \subset X$ for $y \in Y$ is also a connected
submanifold of same dimension. We obtain an exact sequence of
tangent spaces
$$ 0 \to T_{X/Y} \to T_X \to f^*T_Y \to 0 $$
where $T_{X/Y}$ is the kernel of induced map of tangent bundles. intuitively what is really going on there is that for every $x \in f^{-1}(y)$, the $(T_{X/Y})_x$ is the tangent space of the fiber at $x$. The relative space of Kähler differentials
$\Omega_{X/Y}$ defined as the dual of $T_{X/Y}$ and sits in the sequence which we will
obtain if we dualize the sequence above of tangent spaces:
$$ 0 \to f^*\Omega_Y \to \Omega_X \to \Omega_{X/Y} \to 0 $$
Another definition of relative Kähler differentials which is more common to use in modern algebraic geometry works as follows:
Let us embedd $X$ as the image of the diagonal $\Delta: X \to X \times_Y X$
and assume that the ideal sheaf $I \subset O_{X \times_Y X}$ defines the closed image
$\Delta(X) \subset X \times_Y X$.
This gives us another sequence of tagent spaces
$$ 0 \to T_{\Delta(X)} \to T_{X \times_Y X} \to N_{X \times_Y X/X} \to 0 $$
with normal bundle $N_{X \times_Y X/X}$. It's a basic fact that the dual
of $N_{X \times_Y X/X}$ is $I/I^2$ and most books on algebraic
geometry define the sheaf relative Kähler differentials by
$$\Omega_{X/Y} := I/I^2$$
Since this definition not uses that $f$ is a map of smooth maps this
is a far generalization of the old fashion setting from
differential geometry. Now, if there is any justice n this world then these two definitions of relative differentials should coinside if we deal with $X, Y$ and $fY$ nice enough.
Therefore, if $f$ a surjective map between conneted manifolds such that
every fiber is a connected submanifold, why the tangent bundle
$T_{X/Y}$ and the pullback of normal bundle $\Delta^* N_{X \times_Y X/X}$ of
$\Delta(X) \subset X \times_Y X$ are canonically isomorphic?
Can we write down an explicit isomorphism and understand what is geometrically going on there?
Best Answer
Given a linear map $g \colon V \rightarrow W$ between finite dimensional vector spaces, we can form the exact sequence
$$ 0 \rightarrow V \xrightarrow{\Delta} V \times_{W} V \rightarrow (V \times_{W} V )/ \operatorname{im}(\Delta) \rightarrow 0 $$
where $V \times_{W} V = \{ (v,v') \, | \, g(v) = g(v') \}$ and $\Delta(v) = (v,v)$ is the diagonal embedding. Then we can see that $(V \times_{W} V) / \operatorname{im}(\Delta)$ is canonically isomorphic to $\ker g \subseteq V$ via the subtraction isomorphism $[(v,v')] \mapsto v - v'$ (whose inverse is given by $v \mapsto [(v,0)]$).
This applies fiberwise to the situation you describe for manifolds. Let's assume that $f$ is a surjective submersion (this is enough to guarantee that the fibered product actually exists in the category of smooth manifolds). Fix $x \in X$ and pull back the short exact sequence via $\Delta$ to obtain $$ 0 \rightarrow \Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x} \rightarrow \Delta^{*} T(X \times_{Y} Z)|_{x} \rightarrow \Delta^{*} (N_{\operatorname{im}{\Delta} \hookrightarrow X \times_{Y} X})|_{x} \rightarrow 0 $$
Using the identifications $$ T(X \times_{Y} X)|_{(x,x')} = \{ (v,v') \in T_x X \times T_{x'} X, \, | \, df|_{x}(v) = df|_{x'}(v') \}, \\ \Delta^{*} T(X \times_{Y} Z)|_{x} = T_x X \times_{T_{f(x)} Y} T_x X, \\ \Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x} = \{ (v,v) \in T_x X \times T_x X \} \cong T_x X, \\ T_{X/Y}|_{x} = \ker \left( df|_{x} \right) $$ we see that we are in the algebraic situation described in the beginning of the answer (with $V = T_x X, W = T_{f(x)} Y$ and $g = df|_{x}$).
To see what is going on geometrically, it might be useful to start with the case where $Y = \textrm{pt}$ so that the fibered product is just a regular product $X \times X$ and the relative tangent bundle is the regular tangent bundle of $X$. Take $X = \mathbb{R}$ and identify each equivalence class in a fiber of the normal bundle of $X$ inside $X \times X$ with a slanted line. The isomorphism then identities each such slanted line with its intersection with the (say) $x$ axis (which is the "unslanted, regular" tangent space).