// Workers AI · dad joke modeDoes preimage theorem have a map? It's a precursor to a great image.

In mathematics, particularly in the field of differential topology, the preimage theorem is a variation of the implicit function theorem concerning the preimage of particular points in a manifold under the action of a smooth map.

Statement of Theorem

Definition. Let $f:X\to Y$ be a smooth map between manifolds. We say that a point $y\in Y$ is a regular value of $f$ if for all $x\in f^{-1}(y)$ the map $df_{x}:T_{x}X\to T_{y}Y$ is surjective. Here, $T_{x}X$ and $T_{y}Y$ are the tangent spaces of $X$ and $Y$ at the points $x$ and $y.$

Theorem. Let $f:X\to Y$ be a smooth map, and let $y\in Y$ be a regular value of $f.$ Then $f^{-1}(y)$ is a submanifold of $X.$ If $y\in {\text{im}}(f),$ then the codimension of $f^{-1}(y)$ is equal to the dimension of $Y.$ Also, the tangent space of $f^{-1}(y)$ at $x$ is equal to $\ker(df_{x}).$

There is also a complex version of this theorem:

Theorem. Let $X^{n}$ and $Y^{m}$ be two complex manifolds of complex dimensions $n>m.$ Let $g:X\to Y$ be a holomorphic map and let $y\in {\text{im}}(g)$ be such that ${\text{rank}}(dg_{x})=m$ for all $x\in g^{-1}(y).$ Then $g^{-1}(y)$ is a complex submanifold of $X$ of complex dimension $n-m.$

See also

• Fiber (mathematics) – Set of all points in a function's domain that all map to some single given point
• Level set – Subset of a function's domain on which its value is equal