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

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 $$\partial W= X\times \{0\} \cup Z \times \{1\}$$

Boundary Theorem: suppose that $$X$$ is the boundary of some manifold $$W$$ and $$g: X \to Y$$ is a smooth map. If $$g$$ may be extended to all of $$W$$ then $$I_2 (g,Z)=0$$ for any closed submanifold $$Z$$ in $$Y$$ of complementary dimension.

Assume that $$X$$ deformed into $$Z$$ then there is a homotopy $$i_t: X \to Z$$ such that $$i_0 (x)=x$$ adn $$i_t$$ is embedding, meaning for each $$T$$, $$i_t$$ is injective and the preimmage of any compact set it also compact.

I can't see how these information help me to prove this statement.

This is where pictures really make the intuition clear. I visualize a smooth homotopy $i_t \colon X \to Z$ as a "curvy cylinder" where the bottom of the cylinder is $i_0 \cong X$ and the top of the cylinder is $i_1 \cong Z$. Note that I am using $\cong$ to denote topologically isomorphic.

The bottom of this "cylinder manifold" is then clearly $X \times \{0\}$ while the top is $Y \times \{1\}$. So the boundary of this "cylinder manifold" is just the union of the two.

The definitions you use are embedding $X$ and $Z$ in some ambient manifold $Y$, so a "curvy cylinder" between $X$ and $Z$ should be some "curvy cylinder submanifold" of $Y \times I$.

That's the intuition. For the (not so painful) rigor, see the spoilers below:

Let $Y$ be a manifold and $X,Z \subseteq Y$ be submanifolds. Also, $X$ be deformable into $Z$. That is, there exists a smooth homotopy $i \colon X \times [0,1] \to Y$ where $i_0 \equiv i(X,0) = X \times \{0\} \cong X$ and $i_1 \equiv i(X,1) = Z \times \{1\} \cong Z$.

Now define $g \colon X \times I \to Y \times I$ where $g(x,y) = i_y(x) \times \{x\}$. Since each $i_x$ is an embedding, $g$ is also an embedding $W \equiv g(X,Y) \subseteq Y \times I$. Note that via the obvious chart maps, $W$ has a boundary, namely $\partial W = X \times \{0\} \cup Y \times \{1\}$.

Furthermore, since by definition $X$ is a compact manifold, then via the embedding $g$ you can pull back any open covering of $W$ onto $X \times I$ to find a countable subcover, thus $W$ is compact.