# Definition of differential of map (in algebraic geometry)

Given two schemes $X, Y$ with a map $f: X \rightarrow Y$, one should have a map $df: T^*Y \times_Y X \rightarrow T^*X$. How is this map defined (using the language of algebraic geometry, rather than the language of manifolds)?

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One doesn't usually have a cotangent bundle so much as a cotangent sheaf, in which case it is not appropriate to use fibre product notation for pullback. – Zhen Lin Dec 18 '12 at 4:08
And the cotangent sheaves are relative some basis. – user18119 Dec 18 '12 at 13:51

## 1 Answer

The map of schemes $f:X\to Y$ comes with a map $f^\sharp:\mathcal O_Y\to\mathcal f_*O_X,$ which induces a local homomorphism $f^\sharp_x:\mathcal O_{Y,f(x)}\to\mathcal O_{X,x}$ of local rings. In particular, $f^\sharp_x$ induces a homomorphism of Zariski cotangent spaces $\varphi:\mathfrak m_{f(x)}/\mathfrak m_{f(x)}^2\to\mathfrak m_x/\mathfrak m_x^2.$ (The dual to this map is usually denoted $d_xf: T_{X,x}\to T_{Y,f(x)}.$ This differential is defined locally.)

In the case that $X,Y$ are nonsingular varieties, we do have cotangent bundles $\Omega_X^1\to X,\Omega^1_Y\to Y,$ and supposing that $x\in X,y\in Y$ are closed points, the fibres are $\Omega^1_{X,x}=\mathfrak m_x/\mathfrak m_x^2$ and $\Omega^1_{Y,y}=\mathfrak m_y/\mathfrak m_y^2.$ We can pull back the cotangent bundle of $Y$ to $X$ along $f,$ which gives a vector bundle $f^*(\Omega_Y^1)=X\times_Y\Omega_Y^1\to X.$ Since we are working over an algebraically closed field, the closed points of $f^*(\Omega_Y^1)$ correspond to pairs of closed points of $X$ and $\Omega_Y^1.$ Thus, we can define a morphism $f^*(\Omega_Y^1)\to\Omega_X^1$ by specifying its values on the closed points $(x,\eta)\in X\times_Y\Omega_Y^1.$ The natural choice is the map $(x,\eta)\mapsto\varphi(\eta),$ and this choice is well-defined because $\eta\in\Omega^1_{Y,f(x)}$ by definition of fibre product. (The details of what I've written are very nicely laid out in Shafarevich BAG 2)

As others mention in the comments, the fact that we have a vector bundle is a special property, relying here on the fact that $X,Y$ are nonsingular varieties. In general we have relative cotangent sheaves, and the techniques used shift away from this directly geometrical perspective.

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