In Lee‘s Introduction to smooth manifolds he states the Transversality Homotopy Theorem as follows:

Suppose $M,N$ are smooth manifolds and $X \subset M$ is an embedded submanifold. Every smooth map $f:N\to M$ is homotopic to a smooth map $g: N \to M$ that is transverse to $X.$

I think that this is not true without any further conditions, because if $\dim X + \dim N< \dim M,$ then $\dim T_{g(x)}X+\dim dg_x(T_x N)<\dim T_{g(x)}M$ should hold for all smooth maps $g:N\to M$ and all $x\in N,$ i.e. there should not be any map $g:N \to M$ transverse to $X.$

Is this correct or am I missing something? Thanks in advance.

  • $\begingroup$ Furthermore in the proof of Theorem 6.35 I think that equation (6.9) is wrong and that it should be $dF(T_{(p,s)}W)\subset T_qX$ (for example taking two intersecting planes in $\mathbb{R}^3$ should show that only inclusion is possible to be certain), however the proof should still work. $\endgroup$ Mar 21, 2018 at 20:20
  • $\begingroup$ Yes, if $\dim X + \dim N< \dim M$ then $\dim T_{g(x)}X+\dim dg_x(T_x N)<\dim T_{g(x)}M$. That does not in any way imply that no maps are transverse to $X$. $\endgroup$ Mar 21, 2018 at 22:16
  • $\begingroup$ Isn‘t the definition of being transverse that $T_{g(x)}X$ and $dg_x(T_xN)$ span $T_pM$ for all $x \in g^{-1}(X)$? So is the point that every statement about the empty set is true? $\endgroup$ Mar 21, 2018 at 23:32
  • $\begingroup$ Yes. Try an example, e.g. $\dim X=\dim N=0$ and $\dim M=1.$ $\endgroup$
    – Dap
    Mar 22, 2018 at 6:13

1 Answer 1


The theorem is true. It just means that if $\dim(X)+\dim N<\dim M$ that $g$ will not intersect $X$. So two circles in $\mathbb R^3$ in general position will not intersect. The same for a circle and a point in $\mathbb R^2$.


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