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Let $X$ be a 4-manifold and $T$, $S$ two tori embedded in $X$. Let $m_1$, $l_1$ and $m_2$, $l_2$ be loops in $X$ generating $H_1$ of $T$ and $S$, respectively (where I am identifying the tori with their images in $X$). Suppose that $m_1$ and $m_2$, and $l_1$ and $l_2$ are homotopic in X, respectively. Does it follow that $T$ and $S$ represent homologous elements in H_2(X)?

I would like to use that $m_1$, $m_2$, and $l_1$, $l_2$ bound annuli in $X$ (is this correct?) to construct a map $S^1\times S^1\times I \to X$ with $S^1\times S^1\times 0 \to T$, and $S^1\times S^1\times 1\to S$. But I can't see how to do this.

(I had previously mistakenly posted this without specifying the dimension of X.)

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It doesn't follow.

(I'm taking all of this from the bottom of page 19 in Terry Lawson's survey "The Minimal Genus" which can be found here. What follows is almost an exact copy of his argument, but specialized to the example at hand.)

Let $X = S^2\times S^2$. Take $a=[S^2\times \{p\}]$ and $b=[\{p\}\times S^2]$ as generators for $H_2(X)$. Consider the element $2a+2a\in H_2(X)$. I claim that this is represented by an embedded $T^2$.

Believing the for a second, we can find another $T^2$ in $X$ which is not homologous to this first $T^2$, just look at $T^2\rightarrow \mathbb{R}^4\rightarrow X$ where the first map is any embeding and the second map is a chart map. Since this maps factors through $\mathbb{R}^4$, it's null homologous. This implies the two $T^2$s are not homologous to each other. Nevertheless, since $X$ is simply connected, the condition on your generating curves $m_1, l_1$, etc. is satisfied.

Here's how to construct the embedding. Pick two distinct points, $\{p,q\}\subseteq S^2$ and consider $S^2\times \{p\} \cup S^2\times \{q\} \cup \{p\}\times S^2\cup \{q\}\times S^2 := M_p\cup M_q\cup N_p\cup N_q$ which clearly represents $2a + 2b$. Notice that $M_p\cap M_q = N_p\cap N_q = \emptyset$ but we have $M_i \cap N_j = \{(i,j)\}$ for $i \in \{p,q\}$ and likewise for $j$. So, there are 4 intersection points: $\{(p,p), (p,q), (q,p), (q,q)\}$.)

Connect sum $M_p$ and $N_p$ at the point $(p,p)$. Since $M_p$ and $N_p$ were spheres, this connect sum is still a sphere. But now we've cut out the "bad" point and replaced it with an embedded tube.

I claim the new thing represents the same homology class as the original union of the 4 spheres. The point is that the natural map $M_p\sharp N_p\rightarrow M_p\vee N_p$ (the one point union) carries the fundamental class $[M_p\sharp N_p]$ to $[M_p]+[N_p]\in H_2(M_p \vee N_p) \cong H_2(M_p)\oplus H_2(N_p)$. So, at least as homology is concerned, the inclusion map of $M_p\sharp N_p$ into $X$ factors through the map $M_p\sharp N_p\rightarrow M_p\vee N_p$.

Likewise, connect sum at all the other intersection points except $(q,q)$. At this stage one has an immersed sphere representing $2a+2b$. This sphere is not embedded as it has a double point: $(q,q)$. Finally, "connect sum" (i.e., do surgery) there. In this case, the two neighborhoods removed are from the same sphere, so after gluing in the tube, one gets $T^2$ which is now embedded.

$$ $$

Finally, if one wants to do this on $X = \mathbb{C}P^2$ (as per our original discussion), consider the element $1\in H_2(\mathbb{C}P^2)$ represented by, say, the sphere $S^2 = \{[z_0:z_1:0]\in \mathbb{C}P^2\}$. Then the element $3$ is represented by $\{[z_0:z_1:0]\} \cup \{[z_0:0:z_1]\} \cup \{[0:z_1:z_2]\}$. Each of these 3 sets is an $S^2$ and any two of them intersect in a single point (either $[1:0:0], [0:1:0],$ or $[0:0:1]$). One does the same construction as above to get an embedded torus representing $3\in H_2(\mathbb{C}P^2)$.

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I learned a whole lot while trying to adapt my argument from your previous question to this! Thanks for posing the question, it was tons of fun! –  Jason DeVito Dec 12 '12 at 16:27

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