Let $U$, $V$, and $W$ be vector spaces over the same field $k$. Then there are isomorphisms given by \begin{align*} (U\otimes V)\otimes W\cong U\otimes (V\otimes W) \end{align*} given by, \begin{align*} (u\otimes v)\otimes w\mapsto u\otimes (v\otimes w). \end{align*} The difficuly I am having is the following. In order to prove that this is an isomorphism I would need to show that this mapping preserves the vector space structure from one product to the other. My work so far goes as follows.
Let $f$ denote the mapping above. We wish to show that $f(a+b)=f(a)+f(b)$ and $f(\lambda a)=\lambda f(a)$ where $a\in(U\otimes V)\otimes W$ and $b\in U\otimes (V\otimes W)$. For the second identity we have, \begin{align*} f(\lambda a)&=f(\lambda ((u\otimes v)\otimes w))=f((u\otimes v)\otimes \lambda w)\\ &=u\otimes (v\otimes \lambda w)=u\otimes \lambda(v\otimes w)=\lambda(u\otimes(v\otimes w))\\ &=\lambda f(u\otimes(v\otimes w))=\lambda f(a). \end{align*}
For the first identity we have, \begin{align*} f(a+b)=f(((u\otimes v)\otimes w)+((u'\otimes v')\otimes w'))=? \end{align*}
Here I get stuck because I don't have any algebra I can use to work with the two expressions for two arbitrary objects. The only way I can think to make this work is to just suppose that the provided mapping is assumed to extend linearily like this but then it feels like I'm just assuming it all works out. For further context this particular proposition is in Christian Kassel's book on Quantum Groups. It is part of proposition II.1.3. Any help is greatly appreciated.
Note that not all elements of $(U \otimes V) \otimes W$ are simple tensors. This is the problem you are encountering, as your map isn't defined for elements like $(u_1 \otimes v_1) \otimes w_1 + (u_2 \otimes v_2) \otimes w_2$.
You can fix this by mapping each simple tensor as above and then extending the map linearly on all elements. This trivially gets rid of the linearity of the map. However you might need to bother with the well-definedness of the map a bit.