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Let $G=\mathbb{Z}/2\mathbb{Z}$ be $\{\pm1\}$ and let there be two $G$-spaces given: $X=$ The surface of a cylinder including its boundary circles and $S^4$. That means we two G-actions $f_1:G\times X\to X$ and $f_2:G\times S^4 \to S^4$. Imagine that $f_2$ is actually a family of $G$-actions on $S^4$ parametrized by $\mathbb{Z}/5\mathbb{Z}$, and each member is denoted $f_2^j$ where $j\in\mathbb{Z}/5\mathbb{Z}\equiv\{0,1,2,3,4\}$.

For fixed $j$, how do you go about classifying all $G$-maps from $X\to S^4$ up to $G$-equivariant homotopies?

I've been reading about equivariant homotopy but it seems like an overkill for the $G$ and two spaces I have in mind. There must be a more simple approach.

If it helps, I can specify $f_1$ and $f_2$:

The cylinder is given by $X=\{(z,\theta)\in\mathbb{R}^2|z\in\left[0,1\right]\land\theta\in\left[-\pi,\pi\right]\}/\{(z,\pi) \tilde{} (z,-\pi)\}$.

Then $$(-1, (0,\theta)) \stackrel{f_1}{\mapsto}(0,-\theta)$$ $$(-1, (1,\theta)) \stackrel{f_1}{\mapsto}(1,-\theta)$$ $$(-1, (z,\theta)) \stackrel{f_1}{\mapsto}(z,\theta)\,\,\,\forall z\in\left(0,1\right)$$

If $S^4\equiv \{x\in\mathbb{R}^5|\,\,\,x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=1\}$ then $$ \left(-1,\left(x_1, x_2, x_3, x_4, x_5\right)\right)\stackrel{f_2^j}{\mapsto} \left((-1)^{H(j)})x_1, (-1)^{H(j-1)}x_2, (-1)^{H(j-2)}x_3, (-1)^{H(j-3)}x_4, x_5\right) $$ where $H(j) = 1$ if $j\geq1$ and $H(j) = 0$ otherwise.

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    $\begingroup$ Since you can $G$-equivariantly retract the cylinder to a circle, I don't see the point in including the $z$ coordinate. (I also don't see why you've split up the definition of $f_1$ into three lines, instead of just $(-1,(z,\theta))\mapsto (z,-\theta) \forall z$.) $\endgroup$
    – Allen Knutson
    Commented Mar 7, 2015 at 23:30
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    $\begingroup$ Now your $f_1$ isn't continuous. $\endgroup$
    – Allen Knutson
    Commented Mar 8, 2015 at 0:46
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    $\begingroup$ Is $\ X := S^1\times[0;1]$ ? $\endgroup$
    – Włodzimierz Holsztyński
    Commented Mar 8, 2015 at 3:09
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    $\begingroup$ @WłodzimierzHolsztyński, yes, it is, I just wrote the parametrization explicitly to specify $f_1$. AllenKnutson, are you saying X/~ is the Klein-bottle? I don't see that.. I'm also unsure how you get from $G$-maps to quotient spaces. $\endgroup$
    – PPR
    Commented Mar 8, 2015 at 12:16
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    $\begingroup$ (1) Start with a cylinder including its boundary discs, so, $S^2$. Cutting out an open disc and $Z_2$-identifying its boundary circle is the same as gluing in a cross-cap; that's what you're doing at top and bottom of your cylindrical $S^2$. (2) An equivariant map $X\to Y$ induces a map $X/G\to Y/G$. (But not vice versa: consider $X=Y=Z_2$, with two equivariant maps that become the same in the quotient.) $\endgroup$
    – Allen Knutson
    Commented Mar 9, 2015 at 1:53

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