Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Is $Im T$ necessarily a vector space? – Tim kinsella Apr 2 '13 at 10:02

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 $\operatorname{Im}(T)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.