Normal subloop equals kernel of algebra loop homomorphism

Statement
Suppose $$L, M$$ are algebra loops with $$*$$ as the binary operation and $$e_L, e_M$$ the identity elements. Suppose $$\varphi:L \to M$$ is a homomorphism of algebra loops. In other words:

$$\varphi(x * y) = \varphi(x) * \varphi(y); \qquad \varphi(e_L) = e_M$$.

(the second condition is redundant and can be dropped).

Then, the set $$\varphi^{-1}(e_M)$$, is a normal subloop of $$L$$.

Analogous facts in other algebraic structures

 * Normal subgroup equals kernel of homomorphism
 * Ideal equals kernel of homomorphism of Lie rings

Proof that the kernel is a subloop
Given: A homomorphism $$\varphi:L \to M$$.

To prove: $$\varphi^{-1}(e_M)$$ is a subloop of $$L$$.

Proof:


 * 1) It is closed under multiplication: If $$\varphi(x) = e_M$$ and $$\varphi(y) = e_M$$, then $$\varphi(x * y) = e_M * e_M = e_M$$. Thus, $$\varphi^{-1}(e_M)$$ is closed under multiplication.
 * 2) It contains the identity element: $$\varphi(e_L) = e_M$$ by definition/assumption.
 * 3) If $$a * x = b$$, and $$\varphi(a) = \varphi(b) = e$$, then $$\varphi(x) = e$$. Again, by the homomorphism property, we have \varphi(a * x) = \varphi(b)</math., so $$\varphi(a) * \varphi(x) = \varphi(b)$$, so $$e_M * \varphi(x) = e_M$$. Thus, $$\varphi(x) = e_M$$.
 * 4) If $$y * a = b$$, and $$\varphi(a) = \varphi(b) = e$$, then $$\varphi(y) = e$$. Again, by the homomorphims property, we have $$\varphi(y * a) = \varphi(b)$$, so $$\varphi(y) * \varphi(a) = \varphi(b)$$, so $$\varphi(y) * e_M = e_M$$, so $$\varphi(y) = e_M$$.

Proof that the kernel is normal
To prove: if $$N = \varphi^{-1}(e_M)$$, then for all $$a,b \in L$$, $$a * (b * N) = (a * b) * N = (a * N) * b$$.

Proof: We first claim that for any $$a \in M$$, $$\varphi^{-1}(\varphi(a)) = a * N = N * a$$. Suppose $$c \in \varphi^{-1}(\varphi(a))$$. Then, there exists $$d$$ such that $$a * d = c$$. Then, $$\varphi(a) * \varphi(d) = \varphi(c) = \varphi(a)$$, so by cancellation, $$\varphi(d) = e$$, so $$d \in N$$. Thus, $$c \in a * N$$. A similar argument shows that $$c \in N * a $$.

Conversely, for $$n \in N$$, $$\varphi(a * n) = \varphi(a) * \varphi(n) = \varphi(a) * e = \varphi(a)$$, and similarly $$\varphi(n * a) = \varphi(n) * \varphi(a) = e * \varphi(a) = \varphi(a)$$. Thus, $$\varphi^{-1}(\varphi(a)) = a * N = N * a$$.


 * 1) $$a * (b * N) = (a * b) * N$$: If $$x \in a * (b * N)$$, then $$\varphi(x) = \varphi(a) * \varphi(b) = \varphi(a * b)$$. Thus, $$x \in (a * b) * N$$. Conversely, if $$x \in (a * b) * N$$, and $$x = a * y$$, then $$\varphi(x) = \varphi(a) * \varphi(y)$$, so $$\varphi(a * b) = \varphi(a) * \varphi(y)$$, which gives $$\varphi(b) = \varphi(y)$$ by cancellation, forcing $$y \in b * N$$. Thus, $$x \in a * (b * N)$$.
 * 2) $$a * (b * N) = a * (N * b)$$: It suffices to show that $$b * N = N * b$$, which follows from the fact that both are equal to $$\varphi^{-1}(\varphi(b))$$.