Let $M$ be a differentiable manifold. Prove that the tangent space of $\Delta=\{(p,p):p\in M\}\subseteq M\times M$ at a point $(p,p)$ is the diagonal of $T_pM\times T_pM$.
There are many things that I don't understand and make me feel uneasy. First, I'll write the definition of tangent space I know.
Definition: Let $M$ be a differentiable manifold, $p\in M$ and let $\alpha_1,\alpha_2:(-\epsilon,\epsilon)\to M$ two smooth curves in $M$ such that $\alpha_1(0)=\alpha_2(0)=p$. We define the relation $\alpha_1\sim\alpha_2$ iff there exists a chart $(U,\phi)$ with $p\in U$ such that $(\phi\circ\alpha_1)'(0)=(\phi\circ\alpha_2)'(0)$. Then we define the tangent space as the family of equivalence classes: $$T_pM=\{\overline{\alpha}:\alpha:(-\epsilon,\epsilon)\to M,\alpha(0)=p\}.$$
Here are the things I don't get:
1.- What is the differential structure over $\Delta$? We can only talk about its tangent space if it's a differentiable manifold. Could it be simply the restriction of the charts of the space product $M\times M$? Is it important to know what are the charts, though?
2.- In any case, how could it be true that $T_{(p,p)}\Delta=\Delta(T_pM\times T_pM)$? The elements in $T_{(p,p)}\Delta$ are equivalence classes of curves $\overline{\alpha}$ with $\alpha(0)=(p,p)$. The elements in $\Delta(T_pM\times T_pM)$ are pairs $(\overline{\beta},\overline{\beta})$, which are not equivalence classes I think.
Any help would be appreciated.
In response to your points:
Observe that if $M$ is an $n$-manifold, then $M \times M$ is a $2n$-manifold, and $M$ is an $n$-submanifold of $M \times M$. What does this tell you about the structure of $\Delta(M,M)$? (Hint: construct a chart on $M \times M$ — what do coordinates of points in $\Delta(M,M)$ look like?)
By contradiction: would a curve on $M\times M$ passing through a point in $\Delta(M,M)$ remain so if its tangent vector at $p \in \Delta(M,M)$ were not in $\Delta(T_p M, T_p M)$? What happens on a chart?