Let $M$ be a manifold, and let $A$ and $B$ be two submanifolds of $M$ which are diffeomorphic to each other. I'll say $A$ and $B$ are homotopy equivalent in $M$ if there is a $C^1$ function $f:A \times [0,1] \mapsto M$ such that $f(\cdot,0)$ is is the inclusion map for $A$ and $f(\cdot, 1)$ is the inclusion map for $B$ up to a diffeomorphism. My question is, under what circumstances does homotopy equivalence of $A$ and $B$ imply that $A$ and $B$ are cobordant in $M$ in the sense that there is a submanifold with boundary, $C$, obeying $\partial C=A\cup B$? If $A$, $B$, $C$ need only be immersed submanifolds it seems like this might be true without further restriction, since we can define $C$ as the image of $f$, but if we require $A$, $B$, $C$ to be embedded submanifolds this construction will generically produce self-intersections in $C$ and it is not clear to me under what circumstances these can be "deformed away". Also as pointed out in the comments, even in the immersed case $f$ is not necessarily an immersion, so we still need to ask if it can be deformed into one.
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3$\begingroup$ The version of the question for immersions does not have an obvious answer as you suggest it does. You can't always make a map an immersion using general position arguments. $\endgroup$– John PardonAug 6, 2017 at 17:42
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$\begingroup$ You mean because the derivative matrix of $f$ might not have full rank? Or did you have some other obstruction in mind? $\endgroup$– DanielHarlowAug 6, 2017 at 17:45
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1$\begingroup$ A similar question was asked here: mathoverflow.net/questions/27702/… Note that the homotopy $f$ already gives a null-bordism of $A\cup B$, so you just have to decide if $f$ can be made an immersion (for which there is the Smale-Hirsch theory) and then an embedding (for which there are variants of the Whitney trick, as mentioned in Oscar's answer, but these will be highly sensitive to the dimension/codimension). $\endgroup$– Mark GrantAug 6, 2017 at 20:10
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