Show that there exist a compact manifold with boundary $W$ in $Y \times I$ such that $\partial W= X\times \{0\} \cup Z \times \{1\}$

Question

I found this question had been posted by someone, but got no answer, so I hope I will have better luck.

Show that if $X$ may be deformed into $Z$ then $X$ and $Z$ are cobordant.

Deformation definition: deformation of a submanifold $Z$ in $Y$ is a smooth homotopy $i_t:Z\to Y$ where $i_o$ is the inclusion map $Z\to Y$ and each $i_t$ is an embedding.

Cobordant definition : Two compact submanifolds $X$ and $Z$ in $Y$ are cobordant if there exist a compact manifold with boundary $W$ in $Y \times I$ such that $\partial W= X\times \{0\} \cup Z \times \{1\}$

Here is what I got so far

Assume that $X$ can be deformed into $Z$. Then there exists a smooth homotopy $H: X\times I \to Y$ such that

$$H(x,0)= id_x$$ $$H(X,1)=Z$$

and $H(x,t)$ is an embedding function.

Let $F:X\times I \to Y \times I$ be defined by $F(x,t)=(H(x,t),t)$.

Taking the derivative of $F$ I got

$$dF = \left( \begin{array}{ccc} \frac{\partial H}{\partial x} & \frac{\partial H}{\partial t} \\ \frac{\partial t}{\partial x} & \frac{\partial t}{\partial t} \\ \end{array} \right) $$

Note that $H(x,t)$ is an embedding so $\frac{\partial H}{\partial x}$ is injective, so

$$dF = \left( \begin{array}{ccc} \frac{\partial H}{\partial x} & \frac{\partial H}{\partial t} \\ 0 & 1 \\ \end{array} \right) $$

Now I'm stuck.

Answer 1

You know that for a fixed $t_0$, $H(x,t_0)$ will be an embedding of $X$ in $Y$. You don't a priori know that the whole product space $X\times I$ will be embedded in $Y\times I$. And depending on your definition of $F$, it might not be.

That's where your definition of the map $F(x,t)$ and your work showing that $dF$ is of full rank comes in, proving that $F$ is an immersion. Since it is clearly one-to-one you know that $F$ is an embedding. As $W:=im(F) \subset Y\times I$ is the image of an embedding, it is an embedded submanifold. This allows you to conclude that $X$ and $Z$ are cobordant.