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As in the title, with $X$ and $Y$ smooth manifolds. I am filling in the details of a proof here, and I have a question about identifications of tangent vectors/spaces.

Fix $(p,q)\in X\times Y$. Let $i:x\mapsto (x,q)$ and $\pi:(x,y)\mapsto x$. Then, $L=d(i\circ \pi)$ is a projection, so $\text{ker}(L)\oplus \text{im}(L)=T_{(p,q)}(X\times Y).$

Assume first that $X$ and $Y$ are Euclidean spaces, and take a curve $\gamma:(-\epsilon,\epsilon)\to X\times Y$ with $\gamma (0)=(p,q)$. Now, $\gamma (t)=(\gamma_1(t),\gamma_2(t))$ with $\gamma_1(0)=p$. Then,

$d(\iota_X\circ \pi_X)_{(p,q)}\gamma'(0) = \frac{d}{dt}|_{t=0}(\gamma_1(t),q) = (\gamma_1'(0),0)$

so the image of $L$ is isomorphic to $T_pX.$

Simlilarly, since $di$ is injective, $\text{ker}L=\text{ker}\pi$ so for $\gamma$ as above we have

$(d\pi)_{(p,q)}\gamma'(0)=\frac{d}{dt}|_{t=0}\pi\circ\gamma(t)=\gamma_1'(0).$

If this is to be equal to $0$ then $T_pX=0$ and so $\text{ker}L\cong T_qY.$

Now, suppose that $X$ and $Y$ are arbitrary manifolds, so that now the tangent vector is defined by

$d(\iota_X\circ \pi_X)_{(p,q)}\gamma'(0)(f) = \frac{d}{dt}|_{t=0}f(\gamma_1(t),q)$ for smooth $f:X\times Y\to \mathbb R.$

In the above link to the sketch of a proof, the author uses the first blockquote equation for the arbitrary manifolds $X$ and $Y$, (the second is mine) but isn't there a problem with this? He/she seems to be using the same symbol for different objects and/or making some identification. Namely, the equation that should be used is that in the third blockquote, because if not, then the $\gamma'(0)$ on the LHS is a tangent vector in $T_{(p,q)}(X\times Y$), whereas the RHS is not a tangent vector. The derivative does not even make sense unless we are working in some coordinate chart, but even in this case, it is not clear to me what identification(s) we are using. I am not claiming that the result is false, only that there is some explanation that I am missing here. In fact, I have a general problem with this type of machinery one uses in differential geometry (I am self-studying Lee). I understand how the various definitions of the tangent space are equivalent, but I have trouble understanding equations in which there are implicit identifications made.

Matematleta
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  • in proving this isomorphism, what is the definition of tangent vector you're working with? as equivalence classes of curves, or as derivations? – peek-a-boo Aug 28 '19 at 16:39
  • but in any case, if $\mu : I \to M$ is a mapping of an open subset of $\Bbb{R}$ containing the origin, then of course $\mu'(0)$ cannot mean the standard difference quotient's limit. In this case, my best guess is it equals $d\mu_0(1)$, where $d \mu_0 : T_0I \to T_{\mu(0)}M $ is the tangent mapping between tangent spaces; now note that since $I$ is an open subset of $\Bbb{R}$, their tangent spaces can be canonically identified by the differential of the inclusion map; so $T_0I \cong T_0 \Bbb{R} \cong \Bbb{R}$, where the last isomorphism is because $\Bbb{R}$ is a vector space – peek-a-boo Aug 28 '19 at 16:52
  • So, strictly speaking, the $1$ I wrote in $d\mu_0(1)$ above really means the unit tangent vector in $T_0I$ which corresponds to the element $1 \in \Bbb{R}$ under the isomorphism $T_0I \cong \Bbb{R}$. Now, in your question, $\gamma'(0)$ is a tangent vector in $T_{(p,q)}(X \times Y)$, and $(\gamma_1'(0), 0)$ is a tangent vector in $T_pX \times T_qY$ – peek-a-boo Aug 28 '19 at 16:55
  • @peek-a-boo I am using the following definition of $\gamma'(0)\in T_pM$: if $\gamma: I\ni 0\to M$ s.t. $\gamma(0)=p$, and $f:U\ni p\to \mathbb R$ is smooth, and $U\subseteq M$ is an open set on which there is a defined chart, then $\gamma'(0)f=\frac{d}{dt}f\circ \gamma(t)|_{t=0}$. Of course, in Euclidean space, this coincides with the usual (geometric) notion of tangent vector. I tried the same identifications as yours , but could not get them to work out. – Matematleta Aug 28 '19 at 21:06

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