Homotopic equivalence for $S^5$ without three $S^1$ Consider a standard embedding of $S^5$ in $\mathbb R^6$:
$S^5: \; x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 = 1.$
And consider three circles, which are sections of $S^5$ by $x_1 x_2$, $x_3 x_4$, $x_5 x_6$ planes:
$S_1: \; x_1^2 + x_2^2 = 1, x_3 = x_4 = x_5 = x_6 = 0,$
$S_2: \; x_3^2 + x_4^2 = 1, x_1 = x_2 = x_5 = x_6 = 0,$
$S_3: \; x_5^2 + x_6^2 = 1, x_1 = x_2 = x_3 = x_4 = 0.$
How can I show that $A = S^5 \setminus (S_1 \cup S_2 \cup S_3)$ is homotopy equivalent to $S^3 \vee S^3 \vee S^3 \vee S^4 \vee S^4$?
I thought about homeomorphism between $A$ and $\mathbb R^5 \setminus (\mathbb R \cup S^1 \cup S^1)$ and that maybe it's possible to build a homotopy to some strand of "independent" pieces and to show that they are homotopic to the components of that wedge sum. One one hand, it seems to me that a line and circles in $\mathbb R^5$ are not linked. But on the other hand, it's strange, because then $A$ would be homotopy equivalent to $S^1 \vee (S^2 \vee S^1) \vee (S^2 \vee S^1)$, isn't it?
 A: First of all notice that $1-$dimensional manifolds generically embedded in $n-$manifolds are unlinked for $n\geq4$.
As you already noticed your space $A := S^5 \setminus (S_1 \cup S_2 \cup S_3)$ is homeomorphic to $\mathbb{R}^5 \setminus (\mathbb{R} \cup S_2 \cup S_3)$. This space is homotopically equivalent to $(\mathbb{R}^5 \setminus \mathbb{R}) \vee (\mathbb{R}^5 \setminus S^1)\vee (\mathbb{R}^5 \setminus S^1)$ (since you can move freely the embedded $S^1$s).
Now consider $\mathbb{R}^5 \setminus S^1$: this is the same as considering $S^5 \setminus (S^1\cup p)$ where $p$ is a point, which is homeomorphic to 
$\mathbb{R}^5 \setminus (\mathbb{R}\cup p)$, which is homotopically equivalent to $(\mathbb{R}^5 \setminus\mathbb{R})\vee (\mathbb{R}^5\setminus p)$.
Hence $A$ is homotopically equivalent to:
$$(\mathbb{R}^5 \setminus\mathbb{R})\vee (\mathbb{R}^5 \setminus\mathbb{R})\vee (\mathbb{R}^5\setminus p)\vee (\mathbb{R}^5 \setminus\mathbb{R})\vee (\mathbb{R}^5\setminus p)$$
Now it is easy to show that $\mathbb{R}^5 \setminus p$ deformation retracts on $S^{4}$ and $\mathbb{R}^5 \setminus\mathbb{R}$ deformation retracts on  $\mathbb{R}^4 \setminus p$ which deformation retracts on $S^3$.
Finally you have that $A$ is homotopically equivalent to  $S^3 \vee S^3 \vee S^3 \vee S^4 \vee S^4$.
