# Are the two ways of interpreting the expression $(|a\rangle\otimes|b\rangle)(\langle c|\otimes\langle d|)(|e\rangle\otimes |f\rangle)$ equivalent?

Reading Nielsen and Chuang, I am under the impression that a linear operator on the tensor product can be written in two ways:

$$$$(\left|a\right> \otimes \left|b\right>)(\left\left\left

As if I take a vector $$\left|e\right> \otimes \left|f\right>$$ from a tensor product and apply the operator, I get in the first case:

\begin{align} (\left|a\right> \otimes \left|b\right>)(\left \otimes \left|f\right>) =& (\left|a\right> \otimes \left|b\right>) \left \left\\ =& \left \left \left|a\right> \otimes \left|b\right> \end{align}

And, in the second case,

\begin{align} (\left|a\right>\left\left \otimes \left|f\right>) =& \left|a\right>\left \otimes \left|b\right>\left \\ =& \left \left \left|a\right> \otimes \left|b\right> \end{align}

In the first case, I consider an operator on the tensor product, and in the second case, a tensor product of operators. Are the two equivalent ? All state vectors considered are arbitrary.

Indeed, they are equivalent. The relation you start with is a special case of the following identity: $$$$(A \otimes B)(C \otimes D) = (AC) \otimes (BD)$$$$ where $$A,B,C,D$$ are all operators in some Hilbert space with a tensor product structure $$\mathcal{H} = \mathcal{H}_1\otimes\mathcal{H}_2$$, and $$A,C$$ are operators that belong to $$\mathcal{H}_1$$, and $$B,D$$ to $$\mathcal{H}_2$$. Operators that act on different Hilbert spaces commute, i.e. $$[M_1 \otimes I, I \otimes M_2]=0$$, because anything you do to objects in $$\mathcal{H}_1$$ cannot affect $$\mathcal{H}_2$$ and vice versa. This is why you are allowed to "pass" $$C$$ to the left, through $$B$$, and combine the operators in "like" Hilbert spaces.
Yes they are equivalent, as you correctly computed. You are essentially just observing that $$|a\rangle\!\langle c|d\rangle$$ can be understood either as an inner product ($$\langle c|d\rangle$$) multiplied scalarly with a vector ($$|a\rangle$$), or as an outer product ($$|a\rangle\!\langle c|$$) applied to a vector ($$|d\rangle$$).
A more formal way to observe that this is not completely trivial might be using a more standard algebraic notation. Let $$V$$ be a Hilbert space, and $$u,v,w\in V$$ arbitrary vectors. Denote with $$v^*\in V^*$$ elements of the dual of $$V$$ (i.e. $$v^*:V\to\mathbb C$$ is the linear functional such that $$v^*(v)=1$$ and $$v^*(v_\perp)=0$$ if $$\langle v,v_\perp\rangle=0$$). Denote with $$uv^*$$ the outer product between $$u$$ and $$v$$. You can understand this as the linear operator $$(uv^*): V\ni x\mapsto u \langle v,x\rangle \in V$$.
Then the observation is that (I'll be pedantic with the parentheses to clarify the order of operations): $$(uv^*)(w) = u \cdot (v^*(w)).$$ Note that in the RHS $$v^*(w)\in\mathbb C$$ and thus $$\cdot$$ denotes a scalar product in $$V$$, whereas in the LHS we have the operator $$(uv^*)$$ acting on its input $$w$$.