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Suppose that $X_1$ is an $n_1$-dimensional submanifold of $\mathbb{R}^{N_1}$, and $X_2$ is an $n_2$-dimensional submanifold of $\mathbb{R}^{N_2}$, and let $X=X_1\times X_2$. Let $p_1\in X_1$ and $p_2\in X_2$. Describe the tangent space $T_pX$ in terms of the tangent spaces $T_{p_1}X_1$ and $T_{p_2}X_2$, where $p=(p_1,p_2)$.

By this question we have that $X$ is an $(n_1+n_2)$-dimensional submanifold of $\mathbb{R}^{N_1+N_2}$. By definition of submanifold, for every $p\in X$, there exists an open set $U\in\mathbb{R}^{n_1+n_2}$ and a neighborhood $V_p$ of $p$ in $\mathbb{R}^{N_1+N_2}$ and a diffeomorphism $\phi: U\rightarrow X\cap V_p$. Let $q=\phi^{-1}(p)$.

The derivative at $q$ is a map $d\phi_q:T_q\mathbb{R}^n\rightarrow T_p\mathbb{R}^N$. Basically, this takes a point $(q,v)$ to $(p,D\phi(q)v)$. and the tangent space $T_pX$ to $X$ at $p$ is the image of this linear map. So this image consists of all $(p,w)$ where $w=D\phi(q)v$ for some $v\in\mathbb{R}^n$.

I am trying to prove that $T_pX=T_{p_1}X_1\oplus T_{p_2}X_2$, but I don't quite see how that will go.

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    $\begingroup$ Hint: $T_p X \cong T_{p_1} X_1 \oplus T_{p_2} X_2$ $\endgroup$ – Sammy Black Nov 9 '13 at 6:57
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From $X$ we have both projections $\pi_i : X \rightarrow X_i$, these induce morphisms $d\pi_i : T_{(p_1,p_2)}X \rightarrow T_{p_i}X_i$ But we also have canonical inclusions $\iota_1 = \mathrm{id} \times {p_2} : X_1 \rightarrow X$ and $\iota_2 = p_1 \times \mathrm{id} : X_2 \rightarrow X$, and induce morphisms: $d\iota_i: T_{p_i}X_i \rightarrow T_{(p_1,p_2)}X$.

Now, we can define the following: $$\begin{align*} \pi: T_{(p_1,p_2)}X &\rightarrow T_{p_1}X_1 \oplus T_{p_2}X_2 \\ v &\mapsto (d\pi_1(v),d\pi_2(v)) \end{align*} $$ And $$\begin{align*} \iota: T_{p_1}X_1 \oplus T_{p_2}X_2& \rightarrow T_{(p_1,p_2)}X \\ (v,w) &\mapsto d\iota_1(v)+d\iota_2(w) \end{align*} $$ Because $\pi_1 \circ \iota_2 = p_1$ constantly, and the analogously for $\pi_2 \circ \iota_1 = p_2$ we have: $$\pi \circ \iota(v,w) = (d\pi_1(d\iota_1(v)+d\iota_2(w)),d\pi_2(d\iota_1(v)+d\iota_2(w)))= (v,w)$$ $$\iota \circ \pi(v) = d \iota_1(d\pi_1(v))+d \iota_2(d\pi_2(v)) = v $$

So $T_{(p_1,p_2)}X \cong T_{p_1}X_1 \oplus T_{p_2}X_2$

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