Semiconjugacy and factor map

Remarks

Some authors use the word semiconjugacy without the surjectivity condition and say factor map only when $h$ is onto. If $h$ is moreover a homeomorphism, it is a topological conjugacy. The relation gives $h \circ f^n = g^n \circ h$, so $h$ maps the orbit of $x$ onto the orbit of $h(x)$, and periodic points of $f$ to periodic points of $g$.

Example: let $\Sigma = \set{0, 1}^{\N}$ with the product topology, let $\sigma$ be the shift $(\sigma x)_i = x_{i+1}$, and let $g(y) = 2y \bmod 1$ on the circle $\R / \Z$. The map $h(x) = \sum_{i \ge 0} x_i 2^{-i-1} \bmod 1$ is continuous and onto, since every number in $[0, 1]$ has a binary expansion. One has $h(\sigma x) = \sum_{i \ge 0} x_{i+1} 2^{-i-1} = 2 h(x) - x_0$, which equals $2 h(x)$ modulo $1$, so $h \circ \sigma = g \circ h$. This $h$ is not injective, since $(1, 0, 0, \dots)$ and $(0, 1, 1, \dots)$ have the same image.

Non-example: the inclusion $h \colon \set{0} \to \R$ satisfies $h \circ f = g \circ h$ for $f = \id$ and $g(y) = y/2$, but it is not surjective, so it is not a factor map.