Good space, if possible wedge sum of spheres, which is homotopy equivalent to $\mathbb{C}^3 \setminus \{\text{axes}\}$ When i where thinking of $\mathbb{R}^3$ without axis, i find out (using deformation retraction on unit sphere) that it's homotopy equivalent to wedge sum of five unit spheres, namely
$$
\mathbb{R}^3 \setminus \{\text{axes}\} \sim S^1 \vee S^1 \vee S^1 \vee S^1 \vee S^1
$$
After that we can ask a question, what is $\mathbb{C}^3 \setminus \{\text{axes}\}$ ? I couldn't solve this using the same technique. The problem is that
$$
\mathbb{C}^3 \setminus \{\text{axes}\} \sim \mathbb{R}^6 \setminus\{\text{3 planes, which intersect trivially}\}
$$
$$
\mathbb{R}^6 \setminus\{\text{3 planes, which intersect trivially}\} \sim S^5 \setminus \{S^1, S^1, S^1\}
$$
By the way, i have doubts that last equivalence is true. Even if it's true, its a dead end, and i don't now ho to move forward from this.
 A: Your claim about being equivalent to $S^5$ minus the complement of three embedded circles is correct (with the proof you suggested). In fact, every embedding of three circles into $S^5$ is equivalent (isotopic) to any other, so any three circles will do; pardon me for not sketching a proof.
There is a nice result from transversality theory which says that if M is a manifold and N a closed (in the sense that its complement is open) submanifold of codimension q, the map $\pi_i(M \setminus N) \to \pi_i(M)$ is an isomorphism for $i < q-1$ and a surjection for $i = q-1$. It follows that your space is 2-connected.
Alexander duality implies that your space has $H_3(X;\Bbb Z) = \Bbb Z^3$ and $H_4(X; \Bbb Z) = \Bbb Z^2$. So the answer cannot be so simple as the case of $\Bbb R^3$.
It is not at all obvious, but it is true, that the resulting space is equivalent to $S^3 \vee S^3 \vee S^3 \vee S^4 \vee S^4$. I do not see an easy proof. My sincere apologies for the detail to follow.
I would argue by saying that this space is equivalent to a CW complex with a cell structure with one 0-cell, three 3-cells, and two 4-cells, and trivial cellular differential (any simply connected space is equivalent to a cell complex with a few cells as its homology demands; this is in some appendix in Hatcher's textbook on algebraic topology).
Thus the 3-skeleton is a wedge of 3-spheres. That the 4-cells are attached trivially (up to homotopy) comes from the fact that the cellular differential is zero and the Hurewicz map $\pi_3(S^3 \vee S^3 \vee S^3) \cong H_3(S^3 \vee S^3 \vee S^3)$ is an isomorphism, so triviality of attaching maps can be determined at the level of homology.
