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I understand that there is a universal property which tells me that given a Clifford algebra $Cl(V, q)$, for a linear map $f: V \to A$ ($A$ any associative algebra) satisfying $f(v)f(v) = Q(v)$, $f$ may be extended to a homomorphism between algebras $Cl(V) \to A$.

I am wondering when $A$ happens to be a Clifford algebra. Do all homomorphisms between $Cl(V)$ and $A$ arise this way? My guess is not... If not, is there any general way one may check any given linear map between Clifford algebra is a homomorphism?

Thank you!

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All algebra homomorphisms arise this way, not only when $A$ itself is Clifford, but generally. This is simply because you can restrict any algebra homomorphism from $\mathrm{Cl}(V)$ to $V$, and then the universal property (in which you have forgotten the word "unique") yields that the extension of this restriction to $\mathrm{Cl}(V)$ is your original algebra homomorphism. – darij grinberg Apr 18 '13 at 16:34
The universal property doesn't give you a good way to check whether a given map is an algebra homomorphism (unless it's constructed using the property to begin with). – darij grinberg Apr 18 '13 at 16:35

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