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I was reading the proof of Lemma 1.25 in this thesis and I thought I understood it, but I think I don't. The thing that I don't see clearly is in page 26 where he is showing that $\textrm{ker}\ \eta\subseteq Q$. Why did he only shows it for elements of the form $m\otimes\varphi$?

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  • $\begingroup$ That does look strange, as usually that would not nearly suffice. $\endgroup$ – Tobias Kildetoft Feb 26 '16 at 11:13
  • $\begingroup$ I'm not saying that the proof is wrong, I mean the result of the lemma certainly is true. In fact, the inverse of the homomorphism is pretty obvious. I just don't understand the proof he gives for the contention of the kernel in $Q$. $\endgroup$ – Antonio Feb 26 '16 at 12:30
  • $\begingroup$ Right, I am not sure if his argument is quite complete, but I did not read it through that thoroughly. $\endgroup$ – Tobias Kildetoft Feb 26 '16 at 12:32
  • $\begingroup$ Well, I think I was misunderstanding the proof by Martin, or at least how did he conclude that $\textrm{ker }\eta\subset Q$. He only did it for elements of the form $m\otimes\varphi\in\textrm{ker }\eta$ since the general case of an element $x\in\textrm{ker }\eta$, which is written as a sum $$x=\sum_{i=1}^{n}m_{i}\otimes\varphi_{i},$$ where not necessarily the elements $m_{i}\otimes\varphi_{i}\in\textrm{ker }\eta$, follows the same lines as the case he did. $\endgroup$ – Antonio Feb 29 '16 at 9:39
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I'll take a stab at it and say it is because of the definition of a homomorphism.

For an element $x\in M\otimes N$ we have that $$x=\sum m_i\otimes n_i$$ and for any homomorphism $\varphi:M\otimes N\to Y$ we have that it must be $$\varphi(x)=\varphi(\sum m_i\otimes n_i)=\sum \varphi(m_i\otimes n_i)$$ hence it suffices to show that it is true for any $m\otimes n$ as the rest is just a sum of them.

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  • $\begingroup$ But in general it may be the case that $x$ belongs to the kernel even if some of the $m_{i}\otimes n_{i}$ don't right? $\endgroup$ – Antonio Feb 27 '16 at 17:35
  • $\begingroup$ You can have that in the quotient but we are talking about hte kernel which is a closed subgroup of it, thereofre ALL components of $x$ must be in the kernel and all components can be reduced into that way. $\endgroup$ – Zelos Malum Feb 27 '16 at 17:44
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    $\begingroup$ This answer is incorrect. There is no reason why a kernel should contain those components. For an easy example, consider the natural map from a tensor square to the symmetric square. $\endgroup$ – Tobias Kildetoft Feb 28 '16 at 10:34
  • $\begingroup$ Yes, I was reviewing my Algebra notes and found what you say. I still don't know how to conclude that the kernel is contained in $Q$ and I'm getting worried, I should understand that. $\endgroup$ – Antonio Feb 28 '16 at 11:02

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