Problematic definition of pullbacks of covector fields in Spivak's Calculus on Manifolds This is a follow-up question to a recent one: What is a $k$-form on $[0,1]^k$ in Spivak's Calculus on Manifolds?
In Spivak's Calculus on Manifolds, he defines the integration on chains as follows:


 
where $A$ is some subset of ${\bf R}^n$. 

In this definition, the pullback $c^*\omega$ is very confusing for me. Let me write down how Spivak developed the definitions. 


*

*Suppose $f:{\bf R}^n\to{\bf R}^m$ is a differentiable function and $\omega$ is a $k$-form on ${\bf R}^m$. Then $f^*\omega$ is defined as a $k$-form on ${\bf R}^n$ by
$$
(f^*\omega)_p=f^*(\omega_{f(p)})
$$
which means if $v_1,\cdots,v_k\in{\bf R}^n_p$, then 
$$
(f^*\omega)_p(v_1,\cdots,v_k)=\omega_{f(p)}(df_p(v_1),\cdots, df_p(v_k)).
$$
Here I have changed Spivak's notation $(f^*\omega)(p)$ to $(f^*\omega)_p$ to make the notation "clearer". 

*On the other hand, a singular $k$-cube $c$ in $A\subset{\bf R}^n$ is defined as a continuous function $c:[0,1]^k\to A.$ (In fact, Spivak only defines the $k$-cube for $k=n$). 
Here comes my question: 

The domain of $c$, namely $[0,1]^k$, is a closed subset of ${\bf R}^k$. one cannot even talk about $dc_p$ for $p$ on the boundary of $[0,1]^k$. How should I make sense of the notation $c^*\omega$ in the quoted definition of integration on chains?  

 A: Worse, if we only require that $c : [0, 1]^k \to A$ be continuous (and not differentiable), $dc_p$ may not even be defined for an interior point $p$! We can rectify this situation by asking for $c$ to be differentiable; by this we mean $c$ is the restriction of some differentiable function $\tilde c : \Bbb R^k \to A$. (We can equivalently replace $\Bbb R^k$ in this definition with any open set containing $[0, 1]^k$.)
In this case, if $c$ is differentiable, then we can check that for any $p \in [0, 1]^k$ (in particular for $p$ on the boundary) $d\tilde c_p$ of the choice of extension $\tilde c$ of $c$. Thus, we can unambiguously declare $dc_p = d\tilde c_p$.
Thinking about the $k = 1$ case here is instructive: If a function $f : [0, 1] \to \Bbb R$ is differentiable, then $df_0 = d\tilde f_0 = \tilde f'(0) \,dx$, and $\tilde f'(0)$ coincides a fortiori with the one-sided limit
$$\lim_{h \searrow 0} \frac{\tilde f(h) - \tilde f(0)}{h} .$$
But $\tilde f$ extends $f$, so this is equal to
$$\lim_{h \searrow 0} \frac{f(h) - f(0)}{h} ,$$
which manifestly does not depend on the choice $\tilde f$ of extension.
