Showing product of two transversal maps is a transversal

Question

Let $X_1,X_2,M$ be finite dimensional manifolds. Consider maps $f_i:X_i\to M$ such that $f_1,f_2$ are transversal with respect to each other, i.e, for $x_1\in X_1, \ x_2\in Y_1$ with $f_1(x_1)=f_2(x_2)=m$, we have $T_{x_1}f_1(T_{x_1}X_1)+T_{x_2}(T_{x_2}X_2)=T_{m}(T_{m}M)$. Show that $f_1,f_2$ are transversal if and only if $f_1\times f_2: X_1\times X_2\to M\times M$ is a transversal to $\Delta_M=\{(m_1,m_2)\in M\times M: m_1=m_2\}$.

Answer 1

Remark that $f_1$ is tranversal to $f_2$ if and only if $f(x_1)=f(x_2)$ implies that $dim(df_{x_1}(T_{x_1}X_1)))+dim(df_{x_2}(T_{x_2}X_2)))-dim(df_{x_1}(T_{x_1}X_1))\cap df_{x_2}(T_{x_2}X_2)))=m$ where $m=dimM$.

$df_{(x_1,x_2)}(f_1\times f_2)(T_{x_1}X_1)\times T_{x_2}X_2)\cap T_(m,m)(M\times M)=\{(u,u), u\in df_{x_1}(T_{x_1}X_1))\cap df_{x_2}(T_{x21}X_2))$. This implies that:

$df_{(x_1,x_2)}(f_1\times f_2)(T_{x_1}X_1)\times T_{x_2}X_2)+dimM-dim(df_{(x_1,x_2)}(f_1\times f_2)(T_{x_1}X_1)\times T_{x_2}X_2)\cap T_(m,m)(M\times M))=dim(df_{x_1}(T_{x_1}X_1)))+dim(df_{x_2}(T_{x_2}X_2)))+m-dim(df_{x_1}(T_{x_1}X_1))\cap df_{x_2}(T_{x_2}X_2)))=2m$, which is equivalent to

$dim(df_{x_1}(T_{x_1}X_1)))+dim(df_{x_2}(T_{x_2}X_2)))-dim(df_{x_1}(T_{x_1}X_1))\cap df_{x_2}(T_{x_2}X_2)))=m$.