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I haven't really found a way to compute this. I wanted to use the duality theorem, i.e. if $A,\,B \subset \mathbb{R}^n$, $A \cong B$, then $$ \newcommand{H}[3] {H_{#1}\left(#2,\,#3\right)} \H{k}{\mathbb{R}^n}{\mathbb{R}^n \setminus A} \cong \H{k}{\mathbb{R}^n}{\mathbb{R}^n \setminus B} $$ and use that $$ f(S^2 \vee S^2) \overset{\text{'compact to T2'}}\cong S^2 \vee S^2 \cong \tau_-(S^2) \cup \tau_+(S^2)\;, $$ where $\tau_{\pm}$ indicates translation by $\pm(1,0,0)$. The set on the right side would then clearly show that $\mathbb{R}^3\setminus f(\tau_-S^2 \vee \tau_+S^2)$ has three path-components; the component given by $\mathbb{R}^3 \setminus (\tau_-D^3 \cup \tau_+D^3)$ being homotopic to $R^3\setminus[-2,2] \times [-1,1] \times [-1,1] \simeq \partial ([-2,2] \times [-1,1] \times [-1,1]) \simeq S^2$ and those given by $\mathbb{R}^3\setminus \tau_\pm (D^2)$.

However, I didn't make much progress (I will probably have to use that the direct sum of the path components' homology is the homology of the space itself - the idea alone hasn't helped me much, though (yet)).

My initial idea was to use the LESes of both pairs and connect them; however, I haven't managed to come up with a morphism $(\mathbb{R}^n,\,\mathbb{R}^n \setminus f(S^2 \vee S^2)) \longrightarrow (\mathbb{R}^3,\, \mathbb{R}^3 \setminus (\tau_-S^2 \cup \tau_+S^2))$ to connect them.

Edit: I just had the idea to apply a kind of thickening argument to get a Mayer-Vietoris sequence related to $\mathbb{R}^3$ minus a closed subset of $\tau_\pm D^3$ which contains the wedge point $0$ and is homotopy equivalent to $\mathbb{R}^3\setminus \tau_\pm D^3$. I don't know if I've tried that already, t.b.h.. Does this idea sound better?

Re-edit: The problem would still be to find a connecting homomorphism. Hmm...

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  • $\begingroup$ Could you say something about your map $f$? I assume it is continuous and $f\colon \Bbb{R}^3\to \Bbb{R}^3$. Then the continuous image of compact is compact, hence $f(S^2\vee S^2)$ is compact. But why $f(S^2\vee S^2)\cong S^2\vee S^2$? $\endgroup$ – Daniel Bernoulli Apr 13 '17 at 9:32
  • $\begingroup$ Do you know Alexander duality? $\endgroup$ – PseudoNeo Apr 13 '17 at 10:07
  • $\begingroup$ @DanielBernoulli $f$ is continuous, I omitted this remark, because this question is 'topology only'. The homeomorphy holds, because $S^2 \vee S^2$ is compact, and $im(f) \subset \mathbb{R}^3$ is T2 - sorry, I forgot to explicitly mention this. $\endgroup$ – polynomial_donut Apr 13 '17 at 13:02
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We have $f(S^2 \vee S^2) \cong S^2 \vee S^2$ because of the injective map $f\colon S^2\vee S^2\to f(S^2\vee S^2)$ which is a homeomorphism, since it is a continuous bijection between compact and Hausdorff.

We replace $R^3$ by its Alexandrov compactification $S^3$ and get by Alexander duality

$$\tilde H_i(S^3\setminus (S^2\vee S^2))\cong \tilde H^{n-i-1}(S^2\vee S^2;\Bbb{Z})\cong \tilde H^{n-i-1}(f(S^2\vee S^2);\Bbb{Z})\cong \tilde H_i(S^3\setminus f(S^2\vee S^2))$$

which implies $H_k(\Bbb{R}^3\setminus f(S^2\vee S^2))\cong H_k(\Bbb{R}^3\setminus (S^2\vee S^2))$.

Consider the homotopy equivalence $\Bbb{R}^3\setminus (S^2\vee S^2)\simeq S^2\coprod \{x_1\}\coprod \{x_2\}$ by deformation retraction, where the deformation retraction inside the spheres gives $x_1,x_2\in\Bbb{R}$ and outside the spheres gives a slightly bigger sphere, which is homotopic equivalent to $S^2$. You can imagine it as a sphere with two disjoint points in it.

Now $$H_k(\Bbb{R}^3\setminus f(S^2\vee S^2))\cong H_k(\Bbb{R}^3\setminus (S^2\vee S^2))\cong H_k(S^2\coprod \{x_1\}\coprod \{x_2\})\cong H_k(S^2)\oplus H_k(pt)\oplus H_k(pt)\cong \begin{cases}\Bbb{Z}^3 & \mbox{for }k=0\\ \Bbb{Z} & \mbox{for }k=2\\ 0 & \mbox{else} \end{cases}$$

and by deformation retraction $$H_k(\Bbb{R}^3)\cong H_k(pt)\cong \begin{cases}\Bbb{Z} & \mbox{for }k=0\\ 0 & \mbox{else} \end{cases}$$

For the topological pair $(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))$ we have a long exact sequence

$$...\to H_k(\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\iota_\ast}{\longrightarrow}H_k(\Bbb{R}^3)\overset{j_\ast}{\longrightarrow}H_k(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\delta_k}{\longrightarrow}H_{k-1}(\Bbb{R}^3\setminus (S^2\vee S^2))\to...$$

where $\iota\colon \Bbb{R}^3\setminus (S^2\vee S^2)\hookrightarrow \Bbb{R}^3$ and $j\colon (\Bbb{R}^3,\emptyset)\to(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))$ induce the maps above and $\delta_k$ is functorial with respect to maps of pairs.

The only interesting part is

$$...\to H_3(\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\iota_\ast}{\longrightarrow}H_3(\Bbb{R}^3)\overset{j_\ast}{\longrightarrow}H_3(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\delta_3}{\longrightarrow}H_{2}(\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\iota_\ast}{\longrightarrow}H_2(\Bbb{R}^3)\overset{j_\ast}{\longrightarrow}H_2(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\delta_2}{\longrightarrow}H_{1}(\Bbb{R}^3\setminus (S^2\vee S^2)) \overset{\iota_\ast}{\longrightarrow}H_1(\Bbb{R}^3)\overset{j_\ast}{\longrightarrow}H_1(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\delta_1}{\longrightarrow}H_{0}(\Bbb{R}^3\setminus (S^2\vee S^2)) \overset{\iota_\ast}{\longrightarrow}H_0(\Bbb{R}^3)\overset{j_\ast}{\longrightarrow}H_0(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2)){\rightarrow}0$$

By diagram chasing we get

  • $H_0(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\cong H_0(\Bbb{R}^3)\cong\Bbb{Z}$
  • $H_1(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\delta_1}{\hookrightarrow}H_{0}(\Bbb{R}^3\setminus (S^2\vee S^2))$ is injective and $\mbox{im}(\delta_1)=\ker(\iota_\ast)$. The inclusion map $\iota:\Bbb{R}^3\setminus (S^2\vee S^2)\to\Bbb{R^3}$ induces $\iota_\ast\colon\Bbb{Z}^3\cong H_0(\Bbb{R}^3\setminus (S^2\vee S^2)){\to}H_0(\Bbb{R^3})\cong\Bbb{Z}$. Since $H_0$ measures the number of path-connected components, we have the three path-connected components in $\Bbb R^3∖(S^2\vee S^2)$ and one component in $\Bbb R^3$. Now $(1,0,0)\overset{\iota_\ast}{\mapsto}1$, $(0,1,0)\overset{\iota_\ast}{\mapsto}1$ and $(0,0,1)\overset{\iota_\ast}{\mapsto}1$, because the inclusion maps every one of the three path-connected components to the one in $\Bbb{R}^3$. Because of homomorphism, we have $\iota_\ast((a,b,c))=a+b+c$. Hence $H_1(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\cong\mbox{im}(\delta_1)=\ker(\iota_\ast)=\langle(1,-1,0),(1,0,-1),(0,1,-1)\rangle$
  • $H_2(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\cong H_{2}(\Bbb{R}^3)\cong 0$, because $\mbox{im}(j_\ast)=\ker(\delta_2)=H_2(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))$ and $\ker(j_\ast)=\mbox{im}(i_\ast)=0$.
  • $H_3(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\cong H_{2}(\Bbb{R}^3\setminus (S^2\vee S^2))\cong \Bbb{Z}$
  • $H_k(\Bbb{R}^3,\Bbb{R}^3\setminus (S^2\vee S^2))\cong 0$ for every $k>3$.
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  • $\begingroup$ Thank you for your answer, I can see you put some work into it. Please consider though, that this is a homework, so this is a little bit too detailed. I haven't read through it all, but the second bullet point seems odd... and wrong ($\iota_*$ everywhere? Coordinate changes? The homology groups aren't $\mathbb{R}^3$(?). The part that I couldn't show was $H_k(\Bbb{R}^3\setminus f(S^2\vee S^2))\cong H_k(\Bbb{R}^3\setminus (S^2\vee S^2))$. You just wrote it, but why does that homeomorphy hold? I think this is the hardest part... $\endgroup$ – polynomial_donut Apr 13 '17 at 13:11
  • $\begingroup$ (I'm talking about the bullet point with $\mbox{im}(\delta_1)=\ker(\iota_\ast)$. Now $(1,0,0)\overset{\iota_\ast}{\mapsto}1$, $(0,1,0)\overset{\iota_\ast}{\mapsto}1$) $\endgroup$ – polynomial_donut Apr 13 '17 at 13:12
  • $\begingroup$ OK. To be not too detailed, I give you a hint: Just adapt the proof in math.stackexchange.com/questions/1007009/… to the restriction $f\colon S^2\vee S^2\to f (S^2\vee S^2) $. The second point gives the map by considering, that the three connected components get mapped to the only one connected component in $\Bbb {R}^3$ by $\iota_\ast $. Maybe I will extend the comment in the next few days. $\endgroup$ – Daniel Bernoulli Apr 16 '17 at 7:11
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    $\begingroup$ Sorry! I didn't read your comment carefully, but I thought this problem would be clear, since PseudoNeo wrote about Alexander duality. We replace $\Bbb{R}^3$ by its Alexandrov compactification $S^3$ and get by Alexander duality $\tilde H_i(S^3\setminus (S^2\vee S^2))\cong \tilde H^{n-i-1}(S^2\vee S^2;\Bbb{Z})\cong \tilde H^{n-i-1}(f(S^2\vee S^2);\Bbb{Z})\cong \tilde H_i(S^3\setminus f(S^2\vee S^2))$. $\endgroup$ – Daniel Bernoulli Apr 18 '17 at 9:50
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    $\begingroup$ $\Bbb{Z}^3\cong H_0(\Bbb{R}^3\setminus (S^2\vee S^2))\overset{\iota_\ast}{\to}H_0(\Bbb{R^3})\cong\Bbb{Z}$ is induced by $\iota:\Bbb{R}^3\setminus (S^2\vee S^2)\to\Bbb{R^3}$. Since $H_0$ measures the number of path-connected components, we have the three path-connected components in $\Bbb{R}^3\setminus (S^2\vee S^2)$ and one component in $\Bbb{R}^3$. Considering $\iota$ the three different path-connected components get mapped to the one in $\Bbb{R}^3$, hence $(1,0,0),(0,1,0),(0,0,1)\overset{\iota_\ast}{\mapsto}1$. By homomorphism property we get $(x,y,z)\overset{\iota_\ast}{\mapsto} x+y+z$. $\endgroup$ – Daniel Bernoulli Apr 18 '17 at 10:02

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