Definition:Concatenation (Topology)

Let $$c_1, c_2: \left[{0 \,. \, . \, 1}\right]^n \to X$$ be maps.

Let $$c_1$$ and $$c_2$$ both satisfy the concatenation criteria $$c \left({\partial \left[{0 \, . \, . \, 1}\right]^n}\right) = x_0$$.

Then the concatenation $$c_1 * c_2$$ is defined as :
 * $$\left({c_1 * c_2}\right) \left({t_1, t_2, \ldots, t_n}\right) = \begin{cases}

c_1 \left({2t_1, t_2, \ldots, t_n}\right) & : \mbox{if } t_1 \in \left[{0 \,. \, . \, 1/2}\right] \\ c_2 \left({2t_1-1, t_2, \ldots, t_n}\right) & : \mbox{if } t_1 \in \left[{1/2 \,. \, . \, 1}\right] \end{cases} $$

where $$\left({t_1, \ldots, t_n}\right) \ $$ are coordinates in the n-cube.

This resulting map is continuous, since:
 * $$2 \left({\tfrac{1}{2}}\right) = 1$$ and $$2\left({\tfrac{1}{2}}\right) - 1 = 0$$;
 * anywhere any co-ordinate of $$\hat{t}$$ is either 1 or 0, $$\left({c_1*c_2}\right) \left({\hat{t}}\right) = x_0$$.

The resulting map also clearly satisfies the concatenation criteria itself.