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Let $V,W$ be vector spaces and $T$ the following mapping: $$ \begin{align*} T:V\times W&\to V\otimes W\\ (v,w)&\mapsto v\otimes w \end{align*} $$ Then $(V\otimes W,T)$ Satisfies the universal property of the tensor product, but if $T$ were surjective, then every element of $V\otimes W$ would be of the form $v\otimes w$ which I am aware is not the case. However, $\tilde T:V\times W\to Im(T)$ with $\tilde T(v,w):=T(v,w)$ is bilinear and therefore, there exists a linear mapping $\phi:V\otimes W\to Im(T)$ so that $\phi\circ T = \tilde T$. Let $j:Im(T)\to V\otimes W$ be the inclusion map. We then have $id\circ T=T=j\circ\tilde T=j\circ\phi\circ T$ which implies $id=j\circ\phi$ by uniqueness of the mapping $\phi$. It follows that $j$ is surjective, and hence $Im(T)=V\otimes W$. This means that $T$ has to be surjective... Where is the mistake that is leading to my utter confusion? Thanks!

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Is $Im T$ necessarily a vector space? –  Tim kinsella Apr 2 '13 at 10:02
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The mistake is in assuming $\operatorname{Im}(T)$ to be a linear subspace of $V \otimes W$, which in general it is not. Actually, $V \otimes W$ is spanned by $\operatorname{Im}(T)$.

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