There is no continuous surjective multiplicative map from $M_n(\mathbb H)$ to $\mathbb H$ Let $\mathbb H$ denote the field of quaternions. I would like to prove that there does not exist any function $f:M_n(\mathbb H)\rightarrow \mathbb H$ for $n\geq 2$ that is continous surjective and multiplicative.  
I have been thinking about this problem for a while but I can't find any contradiction assuming that such a function does exist. I tried considering preimages for $1,i,j,k$ and toying with them, I also tried infering the values of some specific matrices (the $\lambda I_n$, the nilpotent matrices, etc...) but I couldn't reach any conclusion. Mostly, I fail to see how to make use of the continuity here.  
Would somebody have a hint as to how to proceed with this problem?
 A: I will work out the case $n = 2$ in detail. The same proof works for general $n$, I just want to save the labor of typing $n$ by $n$ matrices...
Thus assume that $f:\operatorname M_2 = \operatorname M_2(\Bbb H) \rightarrow \Bbb H$ is a surjective multiplicative map.

Lemma 1. Whenever $A\in \operatorname M_2$ is invertible, the image $f(A)$ is also invertible.

Proof: If $A$ is invertible, then multiplication by $A$ is a bijection on $\operatorname M_2$. Hence $f(A)$ cannot be zero, otherwise $f$ is constantly zero.

Lemma 2. The map $f$ restricted to $\operatorname{GL}_2 = \operatorname{GL}_2(\Bbb H)$ gives a group homomorphism from $\operatorname{GL}_2$ to $\Bbb H^\times$. In particular, we have $f\begin{pmatrix} 1 & \\ & 1\end{pmatrix} = 1$.

Proof: This is clear from Lemma 1.

We want to arrive at a contradiction, hence showing that such an $f$ does not exist.

Assumption 3. Without loss of generality, we may assume that $f\begin{pmatrix} & 1\\1 &\end{pmatrix} = 1$.

Note: for general $n$, we have the canonical embedding of the symmetric group $S_n$ into $\operatorname{GL}_n$, and this assumption becomes: $f(\sigma) = 1$ for all $\sigma \in S_n$.
Why we can make this assumption: we have $f(\sigma)^{n!} = f(\sigma^{n!}) = 1$ by Lemma 2, hence by changing $f$ to $f^{n!}$, which is still surjective multiplicative, we may make this assumption.

From now on, we always make Assumption 3.

Lemma 4. We have $f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}$ for any $\lambda \in \Bbb H$.

Proof: This comes from the identity $\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}\begin{pmatrix} & 1\\1 & \end{pmatrix} = \begin{pmatrix} & 1\\1 & \end{pmatrix}\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}$ and Assumption 3.

Lemma 5. We have $f\begin{pmatrix}1 & \\ & z\end{pmatrix} = 1$ for all $z \in \Bbb H^\times$ with $|z| = 1$.

Proof: For any $\lambda, \mu \in \Bbb H^\times$, we have: $$f\begin{pmatrix}1 & \\ & \lambda\mu\end{pmatrix} = f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}=f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}=f\begin{pmatrix}\lambda & \\ & \mu\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\lambda\end{pmatrix}.$$ Therefore we have $f\begin{pmatrix}1 & \\ & \lambda\mu\lambda^{-1}\mu^{-1}\end{pmatrix} = 1$. But any $z \in \Bbb H^\times$ with $|z| = 1$ can be written as $\lambda\mu\lambda^{-1}\mu^{-1}$ for some $\lambda, \mu \in \Bbb H^\times$.

Lemma 6. For any $a\in \Bbb R$, the value $f\begin{pmatrix}a & \\ & a\end{pmatrix}$ is real.

Proof: Since the matrix $A = \begin{pmatrix}a & \\ & a\end{pmatrix}$ is in the center of $\operatorname M_2$, we have $f(A)f(B) = f(AB) = f(BA) = f(B)f(A)$ for all $B\in \operatorname M_2$. The surjectivity of $f$ then implies that $f(A)$ lies in the center of $\Bbb H$, namely $\Bbb R$.

Assumption 7. Without loss of generality, we may assume that $f\begin{pmatrix}1 & \\ & a\end{pmatrix}$ is real for all $a \in \Bbb R$.

Why we can make this assumption: we already have $$\left(f\begin{pmatrix}1 & \\ & a\end{pmatrix}\right)^2 = f\begin{pmatrix}1 & \\ & a\end{pmatrix}f\begin{pmatrix}a & \\ & 1\end{pmatrix} = f\begin{pmatrix}a & \\ & a\end{pmatrix}\in \Bbb R.$$Therefore, by changing $f$ to $f^2$, we may make this assumption (while still keeping all required properties of $f$, including Assumption 3).

From now on, we always make Assumptions 7.

Lemma 8. We have $f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}\in \Bbb R$ for all $\lambda \in \Bbb H$.

Proof: The case $\lambda = 0$ is covered by Assumption 7. For $\lambda \neq 0$, by Lemma 5 and Assumption 7, we have: $f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}1 & \\ & |\lambda|\end{pmatrix}f\begin{pmatrix}1 & \\ & \frac \lambda {|\lambda|}\end{pmatrix}\in \Bbb R$.


Lemma 9. For any $\alpha \in \Bbb H^\times$, we have $f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = f\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}$.

Proof: This comes from the identity $\begin{pmatrix}1 & \\ & \alpha^{-1}\end{pmatrix}\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}\begin{pmatrix}1 & \\ & \alpha\end{pmatrix} = \begin{pmatrix}1 & \alpha\\ & 1\end{pmatrix}$ and the fact that $f\begin{pmatrix}1 & \\ & \alpha\end{pmatrix}$ is a real number, hence is in the center of $\Bbb H$.

Lemma 10. For any $\alpha \in \Bbb H$, we have $f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = 1$.

Proof: Let $h$ be the value of $f\begin{pmatrix}1 & 1 \\ & 1\end{pmatrix}$. By Lemma 9, we have $h = f\begin{pmatrix}1 & 2 \\ & 1\end{pmatrix} = h^2$. By Lemma 1, we get $h = 1$ and Lemma 9 tells us that $f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = 1$ for any $\alpha \in \Bbb H^\times$. The case $\alpha = 0$ is Lemma 2.

Conclusion. The map $f$ takes values in $\Bbb R$ on $\operatorname M_2$, hence is not surjective. We obtain a contradiction.

Proof: Just note that any matrix in $\operatorname M_2$ can be written as a product of matrices of the form $\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix}$, $\begin{pmatrix} & 1 \\1 & \end{pmatrix}$, $\begin{pmatrix}1 &  \\ & \lambda\end{pmatrix}$ with $\alpha, \lambda \in \Bbb H$ (by performing "row and column operations").

Final remarks.


*

*As claimed in the very beginning, the proof adapts without difficulty to general $n$.

*The continuous assumption is not used. All arguments are algebraic.

*Since it's a proof by contradiction, it doesn't show that any multiplicative map from $\operatorname M_n(\Bbb H)$ to $\Bbb H$ has image in $\Bbb R$. But it is true that any group homomorphism from $\operatorname{GL}_n(\Bbb H)$ to $\Bbb C^\times$ must factorize through $\Bbb R^\times_+$, as the abelianization of $\operatorname{GL}_n(\Bbb H)$ is isomorphic to $\Bbb R^\times_+$.
