5 removed bogus argument

I think that the answer is yes and that we don't really need the full Whitney's trick for this since being homologous is a much coarser relation than being isotopic, which is what Whitney's trick gives. So, rather than using the full trick, one can use just a half of it.

Let $M$ be an oriented connected smooth manifold, and let $Z_1,Z_2$ be oriented pseudo-manifolds representing two cohomology classes (recall that a pseudo-manifold is a stratified space that has no codimension 1 strata and such that each connected component is the closure of a single connected codimension 0 stratum; any homology class can be represented by a pseudo-manifold). Assume $\dim Z_2\geq 1$ [upd: and $\mathop{\mathrm{codim}}_M Z_2>1$; this assumption excludes the cases when one of the cycles is 0-dimensional and the other is of the maximal dimension, in which case the statement we're after is clearly true, and when $\dim Z_1=\dim Z_2=1$; this case has to be considered separately].

First, let's make $Z_2$ connected by joining the connected components with tubes. [upd: as Bruno Martelli points out in the comments, some care is needed here. However, if we have a tube that induces the wrong orientations of one of the components it's supposed to connect, we can always twist the tube since we assume $\mathop{\mathrm{codim}}_M Z_2>1$.] While doing this we may introduce new intersection points, but after a small isotopy these will be all transversal and their signs will add up to 0. Second, take two intersection points $P,Q$ with opposite signs and join them with a non-self-intersecting path $\gamma\subset Z_2$ that does not pass through the singularities [upd: and through other intersecrtion points; some care is needed here as well when $\dim Z_2=1$: in this case we take $P$ and $Q$ to be neighbors on $Z_2$].

Now equip $M$ with a Riemannian metric and let's modify $Z_1$ by taking out two small balls around $P$ and $Q$ in $Z_1$ and inserting a thin tube $T$ instead where $T$ is obtained by exponentiating the sphere subbundle of $N_M Z_2|\gamma$ of sufficiently small radius. More precisely, some work is needed to identify the spheres in $N_MZ_2$ at $P$ and $Q$ with the boundaries of the balls, but this should be no problem.

The result will be homologous to $Z_1$: the fact that $P$ and $Q$ have different signs ensures that the small balls around them and the tube together form the boundary of the exponential of a a ball subbundle of $N_{M}Z_2|\gamma$. [I wish I could draw a picture here but don't know how to do that.] Notice that when $Z_1$ is a loop around $(0,0)$ and $(1,0)$ in $\mathbb{R}^2$ and $Z_2$ is a loop around $(-1,0)$ and $(0,0)$, as in Simon Rose's example, then this procedure cuts $Z_1$ into two loops, one around $(0,0)$, the other around $(1,0)$.

In this way one can eliminate every pair of intersection points with opposite signs. Notice that we haven't done anything to $Z_2$ in the process, apart from making it connected.

[upd: in the case when $M$ is a surface and $Z_1,Z_2$ are 1-dimensional, there are two things one could do. First, one can notice that when $M=S^2$ there is nothing to prove, when $M=T^2$ one can check the statement by hand, and when $M$ is closed and hyperbolic every element of $\pi_1$ (and hence of $H_1$) can be represented by a simple closed loop. Second, then one could try to adapt the above argument for possibly self-intersecting $Z_2$, but this would require some work. In a sense, this makes sense: if 4 dimensions are not enough to perform Whitney's trick, then it's not too surprising that 2 dimensions are not enough to perform half of it.]

4 medium-sized fix

Let $M$ be an oriented connected smooth manifold, and let $Z_1,Z_2$ be oriented pseudo-manifolds representing two cohomology classes (recall that a pseudo-manifold is a stratified space that has no codimension 1 strata and such that each connected component is the closure of its a single connected codimension 0 strata and has no codimension 1 stratastratum; any homology class can be represented by a pseudo-manifold). Assume $\dim Z_2\geq 1$ (when both cycles [upd: and hence $M$ itself are \mathop{\mathrm{codim}}_M Z_2>1$; this assumption excludes the cases when one of the cycles is 0-dimensional and the other is of the maximal dimension, in which case the statement we're after is clearly true)true, and when$\dim Z_1=\dim Z_2=1$; this case has to be considered separately]. First, let's make$Z_2$connected by joining the connected components with tubes. [upd: as Bruno Martelli points out in the comments, some care is needed here. However, if we have a tube that induces the wrong orientations of one of the components it's supposed to connect, we can always twist the tube since we assume$\mathop{\mathrm{codim}}_M Z_2>1$.] While doing this we may introduce new intersection points, but after a small isotopy these will be all transversal and their signs will add up to 0. Second, take two intersection points$P,Q$with opposite signs and join them with a non-self-intersecting path$\gamma\subset Z_2$that does not pass through the singularities .[upd: and through other intersecrtion points; some care is needed here as well when$\dim Z_2=1$: in this case we take$P$and$Q$to be neighbors on$Z_2$]. In this way one can eliminate every pair of intersection points with opposite signs. Notice that we haven't done anything to$Z_2$in the process, apart from making it connected. [upd: in the case when$M$is a surface and$Z_1,Z_2$are 1-dimensional, there are two things one could do. First, one can notice that when$M=S^2$there is nothing to prove, when$M=T^2$one can check the statement by hand, and when$M$is closed and hyperbolic every element of$\pi_1$(and hence of$H_1$) can be represented by a simple closed loop. Second, one could try to adapt the above argument for possibly self-intersecting$Z_2$, but this would require some work. In a sense, this makes sense: if 4 dimensions are not enough to perform Whitney's trick, then it's not too surprising that 2 dimensions are not enough to perform half of it.] 3 a few minor improvements and clarifications; typo fix I think that the answer is yes and that we don't really need the full Whitney's trick for this since being homologous is a much coarser relation than being isotopic, which is what Whitney's trick gives. So, rather than using the full trick, one can use just a half of it. Let$M$be an oriented connected smooth manifold, and let$Z_1,Z_2$be cycles represented by oriented pseudo-manifolds representing two cohomology classes (recall that a pseudo-manifold is a stratified space that is the closure of its codimension 0 strata and has no codimension 1 strata; any homology class can be represented by a pseudo-manifold). Assume$\deg \dim Z_2\geq 1$(when one of the both cycles is and hence$M$itself are 0-dimensional the statement we're after is clearly true). First, let's make$Z_2$connected by joining the connected components with tubes. While doing this we may introduce new intersection points, but after a small isotopy these will be all transversal and their signs will add up to 0. Second, take two intersection points$P,Q$with different opposite signs and join them with a non-self-intersecting path$\gamma$in$\gamma\subset Z_2$that does not pass through the singularities. Now equip$M$with a Riemannian metric and let's modify$Z_1$by taking out two small balls around$P$and$Q$in$Z_1$and inserting a thin tube$T$instead where$T$is obtained by exponentiating the sphere subbundle of$N_M Z_2|\gamma$of sufficiently small radius. More precisely, some work is needed to identify the spheres in$N_MZ_2$restricted to at$P$and$Q$with the boundaries of the balls, but this should be no problem. The result will be homologous to$Z_1$: the fact that$P$and$Q$have different signs ensures that the small balls around them and the tube together form the boundary of the exponential of a a ball subbundle of$N_{M}Z_2$. N_{M}Z_2|\gamma$. [I wish I could draw a picture here but don't know how to do that.] Notice that when $Z_1$ is a loop around $(0,0)$ and $(1,0)$ in $\mathbb{R}^2$ and $Z_2$ is a loop around $(-1,0)$ and $(0,0)$, as in Simon Rose's example, then this procedure cuts $Z_1$ into two loops, one around $(0,0)$ and (0,0)$, the other around$(1,0)$. In this way one can eliminate every pair of intersection points with opposite signs. Notice that we haven't done anything to$Z_2\$ in the process, apart from making it connected.