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I have read "let $f:X \rightarrow Y$ be a $C^{\infty}$ manifold without a boundary and let $Z \subset Y$ be a submanifold of $Y$. Geometrically, we say that $f$ is transverse to $Z$ if $X \times Z$ and $Gr(f) = \{(x, y) \in X \times Y |f(x) = y \}$ are in general position." What does it mean?

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    Wherever you read the above expression, did they earlier define *general position*? Usually transversality means that if $x \in f^{-1}(Z)$ then the image of $Df_x : T_x X \to T_{f(x)} Y$ together with $T_{f(x)}Z$ span $T_{f(x)} Y$.2011-03-27

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Although this certainly isn't quite right, here is what I visualize in my mind when I hear the word transverse. The map $f$ has a differential $df_x$ at every point. If we only consider points on $X$ that map into $Z$, say $p\in f^{-1}(q)$, then $df_p: T_pX\to T_qY$.

In some sense $\mathrm{Im}(df_p)$ is measuring the "direction" of $f$ in $Y$. $Z$ also has a "direction" in $Y$ and we could think of this as $T_qZ\subset T_qY$. We say $f$ is transverse to $Z$ if the collection of all (span of) these directions is the full tangent space at $q$, or $\mathrm{Span}(im(df_p), T_qZ)=T_qY$

For example, take the map $f: \mathbb{R}\to \mathbb{R}^2$ given by $f(x)=(x,0)$. Take as a submanifold of $\mathbb{R}^2$ the $y$-axis. Then the only place where the image of $f$ (which is the $x$-axis) intersects the submanifold is at $(0,0)$. So we only have one point to check, but $im(df_0)=\frac{\partial}{\partial x}$ and $T_{(0,0)}Z=\frac{\partial}{\partial y}$, so the span of these two is all of $T_{(0,0)}\mathbb{R}^2$.

What we see here is that the image of $f$ is the $x$-axis and it intersects the submanifold (the $y$-axis) in a nice way. Note that the $y$-axis could have been any something like the line $y=x$ and it still would be transverse since we are taking span. It didn't need to be perpendicular. If the submanifold was $y=x^2$ then since the intersection point is tangent, we get an example of something that is not transverse.

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    Another test for transversality of submanifolds M,N of X, is that the dimension of the intersection of M and N in X equals the dimension of X minus the sum of the respective codimensions of M,N, i.e., Dim(M/\N)=DimX-Codim(N)-codim(M)2011-04-04