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Let $T: E_1\longrightarrow E_2$ be a bounded linear operator and $T\otimes I:E_1\otimes \mathbb{K}\longrightarrow E_2\otimes \mathbb{K}$ be a tensor product operator , where $E_1, E_2$ be Banach spaces and $I$ is the identity map on $\mathbb{K}$ and $\mathbb{K}$ be scalar field. Does $T\otimes I=\:T$?

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It is not properly an equality, because they are different objects. But they are "equal" in the same sense that $E_1$ and $E_1\otimes\mathbb K$ are equal: for any $x\in E_1$, $k\in\mathbb K$, you have $$ x\otimes k=kx\otimes 1. $$ Similarly, $$ \sum x_j\otimes k_j=\sum k_j\,x_j\otimes 1=\left(\sum k_jx_j\right)\otimes 1. $$ So there is a very natural identification of $E_1$ with $E_1\otimes\mathbb K$. Under this identification, $T$ is mapped to $T\otimes \text{id}$.

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