Homotopy of groups arises from a homomorphism

Statement
Suppose $$G$$ and $$H$$ are groups, and $$\alpha, \beta, \gamma$$ are maps $$G \to H$$ such that $$(\alpha,\beta,\gamma)$$ is a fact about::homotopy of magmas from $$G$$ to $$H$$. In other words, for all $$g_1,g_2\in G$$, we have:

$$\! \alpha(g_1)\beta(g_2) = \gamma(g_1g_2)$$

Then, there is a fact about::homomorphism of groups $$\varphi:G \to H$$ such that:

$$\! \alpha(g) = \alpha(e)\varphi(g), \beta(g) = \varphi(g)\beta(e), \gamma(g) = \alpha(e)\varphi(g)\beta(e) \ \forall \ g \in G$$

where $$e$$ is the identity element of $$G$$.

In particular, since left and right multiplication maps are bijective, all the maps $$\alpha,\beta,\gamma$$ have the same patterns of fibers as the homomorphism $$\varphi$$, namely, cosets of the kernel of $$\varphi$$.

Corollaries

 * Isotopic groups are isomorphic

Proof
Given: Groups $$G,H$$, a homotopy $$(\alpha,\beta,\gamma):G \to H$$

To prove: There is a homomorphism $$\varphi:G \to H$$ such that:

$$\! \alpha(g) = \alpha(e)\varphi(g), \beta(g) = \varphi(g)\beta(e), \gamma(g) = \alpha(e)\varphi(g)\beta(e) \ \forall \ g \in G$$

Proof: Consider the defining equation of a homotopy of magmas:

$$\! \alpha(g_1)\beta(g_2) = \gamma(g_1g_2)$$

Setting $$g_1 = g, g_2 = e$$ gives:

$$\! \alpha(g)\beta(e) = \gamma(g) \ \forall \ g \in G$$

Setting $$g_1 = e, g_2 = g$$ gives:

$$\! \alpha(e) \beta(g) = \gamma(g) \ \forall \ g \in G$$

Combining the above, we get:

$$\! \alpha(e)\beta(g) = \alpha(g)\beta(e) \ \forall \ g \in G$$

This simplifies to:

$$\! \alpha(e)^{-1}\alpha(g) = \beta(g)\beta(e)^{-1} \ \forall \ g \in G$$

Let $$\varphi(g)$$ be this common value. Then $$\alpha(g) = \alpha(e)\varphi(g)$$ and $$\beta(g) = \varphi(g)\beta(e)$$. Also, from the above relation, we obtain that $$\gamma(g) = \alpha(g)\beta(e) = \alpha(e)\varphi(g)\beta(e)$$.

It remains to show that the map $$\gamma$$ is a homomorphism. For this, consider the original defining relation:

$$\! \alpha(g_1)\beta(g_2) = \gamma(g_1g_2)$$

Writing everything in terms of $$\alpha(e), \beta(e), \varphi(g)$$, we obtain:

$$\! \alpha(e) \varphi(g_1)\varphi(g_2) \beta(e) = \alpha(e)\varphi(g_1g_2)\beta(e) \ \forall \ g_1,g_2 \in G$$

Canceling $$\alpha(e)$$ and $$\beta(e)$$ we obtain:

$$\varphi(g_1)\varphi(g_2) = \varphi(g_1g_2) \ \forall \ g_1,g_2 \in G$$

Thus, $$\varphi$$ is a homomorphism of groups.