PlanetPhysics/Cubically Thin Homotopy 2

Cubically thin homotopy
Let $$u,u'$$ be squares in $$X$$ with common vertices.

between $$u$$ and $$u'$$ is a cube $$U\in R^{\square}_3(X)$$ such that
 * 1) A {\it cubically thin homotopy} $$U:u\equiv^{\square}_T u'$$

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 * $$U$$ is a homotopy between $$u$$ and $$u',$$ "i.e. $\partial^{-}_1 (U)=u,\enskip \partial^{+}_1 (U)=u',$" #
 * $$U$$ is rel. vertices of $$I^2,$$ "i.e. $\partial^{-}_2\partial^{-}_2 (U),\enskip\partial^{-}_2 \partial^{+}_2 (U),\enskip \partial^{+}_2\partial^{-}_2 (U),\enskip\partial^{+}_2 \partial^{+}_2 (U)$ are constant,"  #
 * the faces $$ \partial^{\alpha}_{i} (U) $$ are thin for $$ \alpha = \pm 1, \ i = 1,2 $$.

$$u',$$ denoted $$u\equiv^{\square}_T u'$$ if there is a cubically thin homotopy between $$u$$ and $$u'.$$
 * 1) The square $$u$$ is {\it cubically} $$T$$-{\it equivalent} to

This definition enables one to construct $$\boldsymbol{\rho}^{\square}_2 (X)$$, by defining a relation of cubically thin homotopy on the set $$R^{\square}_2(X)$$ of squares.