Confusion regarding notation of a dual transformation I'm reading Spivak's Calculus on Manifolds and in Chapter 4 he defines the dual transformation (although he doesn't call it that) as follows:

If $f:V \rightarrow W$ is a linear transformation, a linear transformation $f^* : \mathfrak{I}^k (W) \rightarrow \mathfrak{I}^k (V)$ is defined by $f^* T (v_1, \dots, v_k) = T(f(v_1),\dots,f(v_k))$ for $T \in \mathfrak{I}^k (W)$ and $v_1, \dots, v_k \in V$.

Here, $\mathfrak{I}^k (V)$ denotes the set of all $k$-tensors on $V$.
My confusion is with the next statement that he makes:

It is easy to verify that $f^*(S \otimes T) = f^*S \otimes f^*T$.

Here, $S$ would be a $k$-tensor on $W$ and $T$ would be an $l$-tensor on $W$. So, on the L.H.S. $f^*$ denotes the map from $\mathfrak{I}^{k+l} (W)$ to $\mathfrak{I}^{k+l} (V)$. But, on the R.H.S. the first $f^*$ denotes the map from $\mathfrak{I}^k (W)$ to $\mathfrak{I}^k (V)$ and the second $f^*$ denotes the map from $\mathfrak{I}^l (W)$ to $\mathfrak{I}^l (V)$. Is this notation standard, so that the dimension of the space on which $f^*$ acts is to be understood from the context? Since we really have three different functions in this equation, is it still appropriate to say that $f^*$ 'distributes over $\otimes$'?
 A: (Apparently I was working with the dual of your definition when first making this post, so let me try again.)
This is just a straightforward unpicking of the definitions:
Let $S \in \mathfrak{I}^k(W)$ and $T \in \mathfrak{I}^l(W)$. Then the $k+l$-tensor $S \otimes T \in \mathfrak{I}^{k+l}(W)$ is given by
$$
 (S \otimes T)(w_1, \dotsc, w_k, w'_1, \dotsc, w'_l)
 = S(w_1, \dotsc, w_k)  \cdot T(w'_1, \dotsc, w'_l).
$$
Thus the $k+l$-tensor $f^*(S \otimes W) \in \mathfrak{I}^{k+l}(V)$ is given by
\begin{align*}
 &\, (f^*(S \otimes T))(v_1, \dotsc, v_k, v'_1, \dotsc, v'_l) \\
=&\, (S \otimes T)(f(v_1), \dotsc, f(v_k), f(v'_1), \dotsc, f(v'_l)) \\
=&\, S(f(v_1), \dotsc, f(v_k)) \cdot T(f(v'_1), \dotsc, f(v'_l)).
\end{align*}
On the other hand $f^* S \in \mathfrak{I}^k(V)$ and $f^* \in \mathfrak{I}^l(V)$ by the way $f^*$ was constructed. So we can form the $k+l$-tensor $(f^* S) \otimes (f^* T) \in \mathfrak{I}^{k+l}(V)$ (notice that this is the same space $f^*(S \otimes T)$ lives in), and this tensor is given by
\begin{align*}
 &\, ((f^* S) \otimes (f^* T))(v_1, \dotsc, v_k, v'_1, \dotsc, v'_l) \\
=&\, (f^* S)(v_1, \dotsc, v_k) \cdot (f^* T)(v'_1, \dotsc, v'_l) \\
=&\, S(f(v_1), \dotsc, f(v_k)) \cdot T(f(v'_1), \dotsc, f(v'_l)).
\end{align*}
By comparing the results of both equations we see that $f^*(S \otimes T) = (f^* S) \otimes (f^* T)$.
PS, regarding the overloading of the notation $f^*$:
If we want to be really rigorous we need to start with denoting the induced map $\mathfrak{I}^k(W) \to \mathfrak{I}^k(V)$ not by $f^*$, but by (something like) $f^*_k$, as the spaces we deal with depend on $k$. So, in a certain way, it is not fine to just write $f^*$.
But you are already doing a lot of overloading before that: For $k,l \geq 0$ and $S \in \mathfrak{I}^k(V)$ and $T \in \mathfrak{I}^l(V)$ you simply write $S \otimes T$, but $\otimes$ is really a map $\mathfrak{I}^k(V) \times \mathfrak{I}^l(V) \to \mathfrak{I}^{k+l}(V)$. So strictly speaking we should also write (something like) $\otimes_{k,l}^V$ instead of just $\otimes$. 
Notice also that using the above, more rigorous notations, the equality $f^*(S \otimes T) = f^* S \otimes f^* T$ is instead written as
$$
 f^*_{k+l}(S \otimes_{k,l}^W T) = f_k^* S \otimes_{k,l}^V f_l^* T,
$$
i.e. the equality states that
$$
f^*_{k+l} \circ \otimes_{k,l}^W
= \otimes_{k,l}^V \circ (f_k^* \times f_l^*).
$$
This notation is much clearer in terms of what’s going on. But when it comes to actually working with this things, we quickly find this notation rather cumbersome and things become pretty unreadable. So just as you are already used to the more sloppy notation of $\otimes$, it is, in a certain way, appropriate to just write $f^*$ instead $f^*_k$ and hope that $k$ is clear from the context. (Which I think most mathematicians do, and understand when they see it somewhere else.
