Homotopy of groups arises from a homomorphism

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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.

Related facts



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.