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I was especially thrown out by the proof on The Transversality Theorem on the point "we want to exhibit a vector $v \in T_x(X)$ such that $df_s(v) - a \in T_z(Z).$"

I understand so far that in order to show that $f_s \pitchfork Z$ we need to show $df_sT_x(v) + T_z(Z) = T_z(Y)$. But why we want to show "a vector $v \in T_x(X)$ such that $df_s(v) - a \in T_z(Z).$"? I guess I lost some general idea here.

I am looking forward a detailed answer which fills up what author omitted.. Thank you!

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So, we need to show $df_s(T_x(X)) + T_z(Z) = T_z(Y)$.

Let $a\in T_z(Y)$ be arbitrary. We want to express it as a sum $$w+w'=a$$ where $w\in df_s(T_x(X))$ and $w'\in T_z(Z)$.
That is, for $w$ there is a $v\in T_x(X)$ such that $w=df_s(v)$, and $w'=a-w$ must be in $T_z(Z)$.

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  • $\begingroup$ Thanks a lot Berci. However, I am still not convinced with the idea that using the behavior of single element ($w+w^\prime= a$) to show the behavior of the set $df_s(T_x(X)) + T_z(Z) = T_z(Y)$. $\endgroup$ – 1LiterTears Jul 13 '13 at 1:08
  • $\begingroup$ @Jellyfish by definition $df_s(T_x(X))+T_z(Z)=T_z(Y)$ implies that for any arbitrary element $z\in T_z(Z)$ can be written as $z=a+b$ where $a\in df_s(T_x(X))$ and $b\in T_z(Z)$. This is just vector space sum. $\endgroup$ – tessellation Jul 22 '13 at 8:50

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