# 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 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 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 $\varphi$.

## 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.