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This sounds intuitively true. However, I have some counter claims:

Although transversal is defined on smooth manifolds, which implies the image of $df_x$ is smooth. But this does not say if the function $f$ itself is smooth, and the existence of $df_x$ only assumes $f$ is $\mathcal{C}^1$.

So can this implies $f$ is smooth, and therefore $f$ generally smooth as long as it transverses?

1LiterTears
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    I don't understand your question - for a smooth map $f:X\to Y$, $df_x$ is a linear map from $T_xX$ to $T_{f(x)}Y$, so of course its image (a linear subspace of $T_{f(x)}Y$) is a smooth submanifold of $T_{f(x)}Y$, and moreover, $df_x$ is only defined under the assumption that $f$ is a smooth map, so how would any of this "imply" that $f$ is a smooth map? – Zev Chonoles Jul 17 '13 at 00:26
  • Oops @ZevChonoles, sorry I failed to express clearly, again. I simply meant to say that as long as I was told that $f$ transverse some space, then I could automatically assume that $f$ is smooth, right~? – 1LiterTears Jul 17 '13 at 00:31
  • Technically, $df_x$ is defined under the assumption that $f$ is $\mathcal{C}^1$, right...? – 1LiterTears Jul 17 '13 at 00:32
  • I suppose that's true (I'll be honest, I haven't ever thought much about $C^k$ manifolds or maps for $k$ finite). – Zev Chonoles Jul 17 '13 at 00:36
  • I edited the question and also responded your comment below @ZevChonoles =) – 1LiterTears Jul 17 '13 at 00:54

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If $f$ is not smooth, then you cannot define $df$ on a local chart. As a result it is impossible to judge the rank of Jacobian and transversality would not make much sense. I suspect you confused whether $df_{x}$ is non-singular with whether $f$ is smooth at $x$.

Bombyx mori
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  • But $df$ is not restrained to be defined on a local chart...? – 1LiterTears Jul 17 '13 at 00:48
  • Also, do you mind explaining why the rank of Jacobian is necessary for transversality? To my understanding, transersality is defined as image$(df_x) + T_x(X) = T_y(Y)$, or $T_x(X) + T_x(Y) = T_x(Z)$. Both are not directly related to Jacobian...? Thank you~.. – 1LiterTears Jul 17 '13 at 05:42
  • You calculate the dimension of $T_{x}(X)$ by the rank of the Jacobian via rank-nullity theorem. Otherwise you do not know what the dimension of the image is like. – Bombyx mori Jul 17 '13 at 05:46
  • Aha, now it makes sense to me. Thank you very much, the great and ancient mathematician! – 1LiterTears Jul 17 '13 at 05:59