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Consider this question from my assignment on manifolds:

Let $M_1$ and $M_2$ be two manifolds and $p\in M_1$ and $q\in M_2$. Prove that $T_{(p,q)}(M_1 \times M_2) $ is isomorphic to $T_p M_1 \times T_q M_2$ as vector spaces.

Attempt: I considered the map $df_{(p,q)}(v) \to (df_p(v), df_q(v))$. I am not able to prove this map to be 1-1. As map df is linear so I assumed that $df_{(p,q)}(v)=(0,0)$, but I am not able to prove that v is 0.

I have proved it onto and vector homomorphism.

Kindly help.

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    Vector sides spaces of equal dimension are always isomorphic, and the dimension of a tangent space is equal to the dimension of the manifold. All you need to show (or know) is that $\dim M_1\times M_2=\dim M_1+\dim M_2$ – Vercassivelaunos Nov 08 '21 at 17:54
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    @Vercassivelaunos Of course, but it is also interesting to consider a certain canonical isomorphism. – spinosarus123 Nov 08 '21 at 18:55
  • For each $(p,q)$ you can define maps $f:M_1\to M_1\times M_2$ and $g:M_2\to M_1\times M_2$ by $f(x)=(x,q)$, $g(y)=(p,y)$. Differentiating these maps gives inculsions $T_pM\to T_{(p,q)}(M_2\times M_2)$, $T_qM\to T_{(p,q)}(M_2\times M_2)$, which are complementary. – Kajelad Nov 09 '21 at 21:25
  • @Vercassivelaunos There certainly can be other methods of proof but can you help proving it by the method used by me? –  Jan 09 '22 at 08:13
  • @spinosarus123 Can you please help with 1-1 part if you have some time to spare? –  Jan 09 '22 at 08:14
  • Which definition of tangent space do you use? What are $f$ and $v$? – Paul Frost Jan 09 '22 at 18:31
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    As Vercassivelaunos comments, it is trivial that there are isomorphic. I think you want to construct a specific isomorphism. – Paul Frost Jan 10 '22 at 11:32
  • @PaulFrost Ya man. –  Jan 13 '22 at 13:57

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Let $pi_1:M_1\times M_2\to M_1$ and $pi_2:M_1\times M_2\to M_2$ be the natural projections. The maps are smooth and we can consider the map $F:T_{(p,q)}(M_1\times M_2)\to T_pM_1\times T_qM_2$, given by $F(v)=(d\pi_1(v),d\pi_2(v))$.

To prove that $F$ is injective let $v$ be a vector such that $F(v)=0$. If we have local coordinates $x_i$ on $M_1$ and $y_i$ on $M_2$ then $x_1,...,x_m,y_1,...,y_n$ are local coordinates on $M_1\times M_2$. It now follows that $$v=\sum a_i\frac{\partial}{\partial x_i}+\sum b_i\frac{\partial}{\partial y_i}.$$ Moreover $$0=d\pi_1(v)=\sum a_i\frac{\partial}{\partial x_i}$$ and $$0=d\pi_2(v)=\sum b_i\frac{\partial}{\partial y_i}$$ so $v=0+0=0$.

spinosarus123
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