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I'm stuck on this question for quite a few days and still haven't got a clue what to do. The question is as follows:

If $\Delta$ is the diagonal of $X\times X$ where $X$ is a manifold, show that its tangent space $T_{(x,x)}(\Delta)$ is the diagonal of $T_x(X)\times T_x(X)$.

Because this question follows a previous part, so I constructed a map

$$f:X\longrightarrow X\times X$$ such that f(x)=(x,x). Therefore I have

$X\overset{f}{\longrightarrow} X\times X$.

Then we take the derivative map

$$T_x(X)\overset{df_x}{\longrightarrow} T_{(x,x)}(X,X)$$

However this does not give me the tangent space of the diagonal...

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    $\begingroup$ Your map $df_x$ is probably not surjective. What could be its range? $\endgroup$ – Thibaut Dumont Jun 7 '13 at 14:09
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    $\begingroup$ Well, I know $f$ is probably not surjective. But I don't know what would happen when taking $df_x$. The diagonal is a manifold itself. So it should have a tangent space at (x,x) as well. Isn't it weird that two manifolds with different dimensions have the same tangent space? Thanks a lot for your help! $\endgroup$ – Evariste Jun 7 '13 at 14:11
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Choose $(v,v) \in \Delta(T_x(X) \times T_x(X)$, with $v \in T_x(X)$. Then there is a smooth curve $\gamma : I \to X$ such that $\gamma(0) = x$, $\gamma'(0) = v$. Now $(\gamma(t) , \gamma(t))$ is a smooth curve that lives in the diagonal of $X \times X$. What happens when you differentiate the curve at $t = 0$? Prove the other direction in the same way.

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    $\begingroup$ Thanks for the nice answer! However I'm not allowed to use this technique. I'm only allowed to do it from scratch(start with a manifold, then take the derivative map. The tangent plane is defined as the image of the derivative map.). Sorry I forgot to mention that in the question. My bad! $\endgroup$ – Evariste Jun 7 '13 at 22:36
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    $\begingroup$ Dear @Evariste I am only using something elementary here: The description of the tangent space as an equivalence class of curves. I suggest you learn about this from say chapter 3 of Lee's Smooth Manifolds, 2nd edition. It is very handy and allows you to compute derivatives of tangent maps very efficiently. $\endgroup$ – user38268 Jun 8 '13 at 0:19
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    $\begingroup$ Hi Ben, I understand what you mean. Curves are more abstract & fundamental because it doesn't depend on a specific coordinate chart. You're right in that sense. But for the purpose of completing this homework question, I can't use it. Of course I'm going to learn ch3 of Lee soon. Thanks for the suggestions. $\endgroup$ – Evariste Jun 8 '13 at 0:29
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    $\begingroup$ Dear @Evariste, I will try to modify my answer to suit your needs. Regards, $\endgroup$ – user38268 Jun 8 '13 at 0:30
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    $\begingroup$ Don't worry about it(for now) Ben. I think I have a proof now. I'll type it up and post it here in about an hour. $\endgroup$ – Evariste Jun 8 '13 at 0:34

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