Let $A$ be a (possibly non-unital) algebra over $\mathbb C$. We say that $A$ is self-induced if the product map $m:A \otimes_A A \rightarrow A$ is an isomorphism. Here $A \otimes_A A$ is the balanced tensor product, the quotient of $A\otimes A$ by the linear span of elements of the form $ab\otimes c – a\otimes bc$.
This notion seems to have been introduced by Gronbaek in Morita equivalence for self-induced Banach algebras and was further studied by Meyer in Smooth and rough modules over self-induced algebras. If $A$ is unital, or more generally, has local, one-sided, units, then $A$ is self-induced. I am interested in non-trivial examples of $A$ which are not self-induced.
What is an example of $A$ which is not self-induced, but such that the product map is surjective, and such that the product is non-degenerate (so for each non-zero $a\in A$ there are $b,c\in A$ with $ba\not=0, ac\not=0$).
I have tagged this functional analysis, as the notion seems to have arisen in the context of topological algebras (and so people working in this area might know examples). But the question does not ask about the topological case. I would be interested to know if this idea is studied in non-topological contexts under a different name?
[Asked on Math.Stackexchange a few days ago, with no answers.]
Partial leads: Some "factorisation" results (relevant to $m$ being surjective in this algebraic situation) are considered by Dales, Feinstein, Pham in Zbl 1471.46051 (full-text available here) which leads me to an old paper of Willis, Zbl 0742.46032. These give very complicated examples of $A$ for which $m$ is surjective, but without there being local units, and so maybe they are not self-induced. Unfortunately, I see no obvious way to check if they are self-induced.
That $m$ is onto means that $A$ is an idempotent ring. Such rings, together with the non-degeneracy condition, and considered by Parvathi and Rao Zbl 0682.16031 and García and Simón Zbl 0747.16007. Unfortunately, these leads do not seem to lead to examples nor consideration of the "self-induced" idea.
Best Answer
In this paper the authors consider the analogous question in the context of semigroups and I think basically the contracted semigroup algebra of their semigroup on page 5 works. This argument should be ok over any base commutative ring with unit but to keep things easier I'll work over a field $K$. If $S$ is a semigroup with zero, then the contracted semigroup algebra $K_0S$ has basis the nonzero elements of $S$ and we interpret a zero product in $S$ as zero in $K_0S$.
Let $S=\{a,b,c,d,0\}$ be the semigroup with zero with the following multiplication table: $$ \begin{array}{c| c c c c c} &0 & a& b &c& d\\\hline 0 & 0& 0& 0& 0& 0\\ a & 0& 0 &0& 0& a\\ b& 0& 0& 0& 0& b\\ c& 0& 0& a& 0& 0\\ d& 0& a& 0& c& d\end{array}$$
Put $A=K_0S$. Note that $S^2=S$ since $d$ is a left identity for $a,c,d,0$ and a right identity for $a,b,d,0$. Therefore, $A^2=A$, i.e., $m\colon A\otimes_A A\to A$ is surjective.
Next I claim that $A$ is non-degenerate. Indeed, if $x=k_aa+k_bb+k_cc+k_dd$ with $xA=0$, then $0=xd = k_aa+k_bb+k_dd$, and so $x=k_cc$. But then $0=xb=k_ca$, hence $x=0$. Similarly, if $Ax=0$, then $0=dx = k_aa+k_cc+k_dd$ and so $x=k_bb$. But then $0=cx = k_ba$ and so $x=0$.
Observe that $A\otimes_{K} A$ has basis all pairs $(s,t)\in S\setminus \{0\}\times S\setminus \{0\}$ and that $A\otimes_A A$ is the quotient of $A\otimes_K A$ by the subspace $V$ spanned by all differences $st\otimes u-s\otimes tu$ with $s,t,u,st,tu\neq S\setminus \{0\}$. Note that $b\notin sS$ for $s\in S\setminus \{b\}$. So a term in the spanning set for $V$ involving an element of the from $b\otimes c$ would have to look like $bt\otimes u-b\otimes tu$ with either $u=c$ or $tu=c$. If $tu\neq c$, then we would need $u=c$ and $bt=b$ to get $b\otimes c$. But then $t=d$ and we get $b\otimes c-b\otimes c=0$. If $tu=c$, then $t=d$, $u=c$. But then we again get $b\otimes c-b\otimes c=0$. Thus $b\otimes c$ does not appear in any term in the spanning set of $V$. Since the $s\otimes t$ with $s,t\in S\setminus \{0\}$ form a basis for $A\otimes_K A$, we must have that $b\otimes c\notin V$. Hence in $A\otimes_A A$, we must have $b\otimes c\neq 0$, but $m(b\otimes c)=bc=0$. So $A$ is our example.
Nb. It looks to me like $S$ is a $\ast$-semigroup with $b^*=c$, $c^*=b$ and all other elements self-adjoint, so you can choose $A$ to be a $\ast$-algebra over $\mathbb C$ if you like and the $\ell_1$-norm will make it a $\ast$-Banach algebra since it is finite dimensional.