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$X, Y$ are vector spaces. A vector space $W$ and a map $A:X\times Y\rightarrow W$ satisfy the following:
for every bilinear map T from $X\times Y$ to a vector space $Z$, there exists a unique linear mapping $L:W\rightarrow Z$ such that $L\circ A=T$. (In fact this is the definition of tensor product)

Then how can I show the image of bilinear mapping $A$ spans $W$?

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If not, let the span be $W'$. There is a linear map $\pi:W\to W$ which has image $W'$ and restricts to the identity on $W'$. Then $I\circ A$ and $\pi\circ A$ are equal, but $\pi\ne I$.

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    $\begingroup$ Oh, so uniqueness is necessary here. Thank you $\endgroup$ – CSH Jul 19 '17 at 6:46

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