why $\overline{f_{q'}} \circ \overline{g_{q}}(q(x))= q'\circ f \circ g(x)$ and $\overline{g_{q}} \circ \overline{f_{q'}}(q(y))= q\circ g \circ f(y)?$

Question

I have some confusion regarding Real Projective space see the page Number 2

It is written that

$$\require{AMScd} \begin{CD} \mathbb{R}^{n+1}\backslash\{0\} @>f>> S^{n} \\ @VqVV @VVq'V \\ \mathbb{RP}^{n} @>>\overline{ f_{q'}}> S^{n}/\sim \end{CD}$$

Let $g: S^{n} \longrightarrow \mathbb{R}^{n+1}\backslash\left\{0\right\}$ and $f: \mathbb{R}^{n+1}\backslash\left\{0\right\} \longrightarrow S^{n}$ defined by $f\left(x\right) = \frac{x}{\left\|x\right\|}$ and $\overline{g_{q}}: S^n/\sim \to \mathbb{RP}^n$ .Then

$\overline{f_{q'}} \circ \overline{g_{q}}(q'(x))= q'\circ f \circ g(x)$

$\overline{g_{q}} \circ \overline{f_{q'}}(q(y))= q\circ g \circ f(y)$

My confusion :Im not getting why $\overline{f_{q'}} \circ \overline{g_{q}}(q(x))= q'\circ f \circ g(x)$ and $\overline{g_{q}} \circ \overline{f_{q'}}(q(y))= q\circ g \circ f(y)?$

My thinking : For $\overline{f_{q'}} \circ \overline{g_{q}}(q'(x))$

we have $\overline{f_{q'}} \circ \overline{g_{q'}}(q'(x)): S^n \to S^n/\sim \to \mathbb{RP}^{n} \to S^n/\sim \implies S^n \to S^n/\sim $

After that I'm not able to proceed further

Answer 1

By definition,

• $\overline{f_{q'}}$ is the map such that $\overline{f_{q'}} \circ q = q' \circ f$; and
• $\overline{g_q}$ is the map such that $\overline{g_q} \circ q' = q \circ g$.

(If this is not clear, read the Lemma 1 again). Thus, $$\overline{f_{q'}} \circ \overline{g_q} \circ q' = \overline{f_{q'}} \circ q \circ g = q' \circ f \circ g, \\ \overline{g_q} \circ \overline{f_{q'}} \circ q = \overline{g_q} \circ q' \circ f = q \circ g \circ f.\, $$