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I am studying geometric measure theory (Herbert Federer - Geometric measure theory) and I have a question about class $r$ homotopies. Here's the definition, from p. 363, Section 4.1.9:

Suppose $U$ is an open subset of $\mathbb{R}^{n}$, $V$ is an open subset of $\mathbb{R}^{v}$ $f$, and $g$ are functions mapping $U$ into $V$. A homotopy of class $r$ from $f$ to $g$ is a map $$ h: A \times U \rightarrow V $$ of class $r$ such that $A$ is an open subinterval of $\mathbb{R}$, $0 \in A$, $1 \in A$, $$ h(0, x)=f(x) \text { and } h(1, x)=g(x) \text { for } x \in U. $$ Whenever $t \in A$ we define $$ h_{t}: U \rightarrow V, \quad h_{t}(x)=h(t, x) \text { for } x \in U, $$ hence $h_{0}=f$ and $h_{1}=g$; in case $r \geq 1$ we also define $$ h_{t}: U \rightarrow \mathbb{R}^{v}, \quad h_{t}(x)=\langle(1,0), D h(t, x)\rangle \text { for } x \in U, $$ hence $\langle(v, w), D h(t, x)\rangle=v h_{t}(x)+\left\langle w, D h_{t}(x)\right\rangle$ for $t \in A$, $x \in U$, $v \in \mathbb{R}$, $w \in \mathbb{R}^{n}$.

Now my question is about the last 3 lines:

in case $r \geq 1$ we also define $$ h_{t}: U \rightarrow \mathbb{R}^{v}, \quad h_{t}(x)=\langle(1,0), D h(t, x)\rangle \text { for } x \in U, $$ hence $\langle(v, w), D h(t, x)\rangle=v h_{t}(x)+\left\langle w, D h_{t}(x)\right\rangle$ for $t \in A, x \in U, v \in \mathbb{R}, w \in \mathbb{R}^{n}$.

Actually, we know that $\langle v, Df(a)\rangle = v \cdot \nabla f(a)$. Also, in the last line it is mentioned that $w \in \mathbb{R}^n$, hence $0 = (0, \dotsc,0) \in \mathbb{R}^n$. Now, we have: $$h_t(x)= \langle (1, 0, \dotsc, 0), Dh(t,x) \rangle$$ which is equal to \begin{align} h_t(x) &= (1, 0, \dotsc, 0) \cdot \nabla h(t,x)\\ &= (1, 0 , \dotsc, 0) \cdot \left(\frac{\partial h}{\partial t}, \frac{\partial h}{\partial x_1}, \dotsc, \frac{\partial h}{\partial x_n}\right) \\ &= \frac{\partial h}{\partial t}.\end{align}

Now, my question is that why it is said that $h_t: U \rightarrow \mathbb{R}^{\nu}$? I mean, now we should see that $\frac{\partial h}{\partial t} \in \mathbb{R}^{\nu}$, but I don't get it really! If this is just the differential of $h$, it must lie in $\mathbb{R}^{\nu-1}$!

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    $\begingroup$ The partial derivative at $t \in A$ of a map $h : A \times U \to V \subseteq \mathbb R^\nu$ is a map $U \to \mathbb R^\nu$; what is the question? (What I don't get is what it means to define $h_t$ twice, on the same domain, in two completely different ways ….) $\endgroup$
    – LSpice
    Commented Jul 31, 2020 at 12:21
  • $\begingroup$ Perhaps the point is that the notation is unclear, and looks like an inner product, whereas I think the significance is that $Dh(t, x)$ is a map $\mathbb R \times \mathbb R^n \to \mathbb R^\nu$, and all we are doing is applying it to the vector $(1, 0) \in \mathbb R \times \mathbb R^n$. (One could think of this as matrix-vector multiplication.) $\endgroup$
    – LSpice
    Commented Jul 31, 2020 at 12:29
  • $\begingroup$ Thank you so much @LSpice. Well, as I saw in the book, the definition of the $\langle -,-\rangle$ is exactly as I wrote and it'll be the inner product of the vector $v=(0,1) \in \mathbb{R} \times \mathbb{R}^n$ and $\nabla h(t,x)$. What I don't understand is that why this inner product lies in $\mathbb{R}^\nu$! I mean, in the definition, it is said that $h$ lies in $\mathbb{R}^\nu$. So, as $h_t = \frac{\partial h}{\partial t}$, shouldn't it lie in $\mathbb{R}^{\nu -1}$? $\endgroup$
    – Zeno cosini
    Commented Jul 31, 2020 at 12:48
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    $\begingroup$ I'm not sure why you would expect the partial derivative of an $\mathbb R^\nu$-valued function to be $\mathbb R^{\nu - 1}$-valued; partial derivatives in some sense "reduce input variables", not "reduce output variables". Anyway, the 'definition' of $\langle{-}, {-}\rangle$ in terms of the gradient that you mention I guess is the one on p. 209, where it is specified only for $Y = \mathbb R$, so that evaluation is the inner product. For more general $Y$, one can only think of evaluation. $\endgroup$
    – LSpice
    Commented Jul 31, 2020 at 13:27
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    $\begingroup$ Yes, you're right @LSpice . I edited the title, thanks for that. About the definition of that product so, how can we define it? I didn't find any definitions and since I'm really new to this topic, I am really confused. $\endgroup$
    – Zeno cosini
    Commented Jul 31, 2020 at 13:52

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