# Thursday March 5th

Theorem (When Cobordisms Are Trivial)
: Let $(W, V, V')$ be a cobordism of dimension $n\geq 6$, $f$ Morse with all critical points of indices in $[2, \cdots, n-2]$.
  Suppose $\pi_1 = 0$ for $W, V,$ and $V'$, and $H_*(W, V) = 0$; then this is homotopic (?) to the product cobordism.

### Proof of Theorem

Factor $c = c_2 c_3 \cdots c_{n-2}$; then from $H_*(W, V) = 0$ we have
\begin{align*}
0 \to C_{n-2} \mapsvia{\del} C_{n-3} \to \cdots \to C_2 \to 0
,\end{align*}
which has zero homology.

Thus for all $\lambda$, choose $\theset{z_1^{\lambda+1}, \cdots, z_{k_{\lambda+1}}^{\lambda+1}  }$ as basis of $\ker C_{\lambda+1} \to C_\lambda$.
Then choose $\theset{b_1^{\lambda+2}, \cdots, b_{\lambda+1}^{\lambda+2} }$ such that $\bd b_i^{\lambda+2} = z_i^{\lambda+1}$.
Then $\theset{b_i^{\lambda}, z_j^\lambda}$ forms an integral basis of $C_\lambda$.

From the basis theorem, we can choose a pair $(f', \xi')$ such that this basis is represented by $D_L$.

![Image](figures/2020-03-18-18:36.png)\

Claim
:   $\bd b^{\lambda+1} = \pm z^\lambda$, so $S_R(q) \intersect S_L(p) = \pm 1$.
    Using the 2nd Cancellation Theorem, the smaller cobordism is thus a product cobordism.

Proof
:   Recall the following:

    - If $X = \bd W$ with $W$ oriented, then $X$ is oriented by $\theset{\nu, \tau_1, \cdots, \tau_{n-1}}$.

    - There is a map

      \begin{align*}
      H_n(W, X) &\to H_{n-1}(X) \\
      [0_W] &\mapsto [0_X]
      .\end{align*}

    So choose an orientation of $W$ (which we'll notate $\circ W$) and all $D_L$, and orient the normal bundle of $D_R$ such that

    - $(\circ (D_R), \circ(D_L) ) = \circ W$
    - $D_R(q_i) \intersect D_L(q_i) = \pm 1$.

    Then $\circ V D_R = \circ VS_R$ and $\circ D_R = \circ S_R$.
    The case for $S_L$ and $VS_L$ are similar.

Lemma
:   Given $(W, V, V')$, let $M \subset V'$ be a smooth submanifold and $[M] \in H_k(M)$ the fundamental class.
    Considering $h: H_k(M) \to H_k(W, V)$, the image

    \begin{align*}
    h([M]) = \sum_{i=1}^\ell \qty{ S_R (q_i) \cdot_V [M] } \cdot D_L(q_i)
    .\end{align*}

Corollary
: If $\bd_{\lambda+1}: C_{\lambda+1} \to C_\lambda$, then
  \begin{align*}
  \bd (q_j') = \sum \qty{ S_R(q_j) \cdot S_L(q_j) } \cdot q_i
  .\end{align*}

This implies the claim.

$\qed$

Proof (of Lemma):
:   Assume $\ell = 1$, then we have a diagram of the form

    \begin{center}
    \begin{tikzcd}
    H_\lambda(M)\ar[drr] \ar[dd] \ar[ddddrrddd, bend right=75] & \\
    &  & H_\lambda(V',V'\setminus S_R)\ar[dd, "h_1"]  \\
    H_\lambda(V') \ar[urr, "h_0"]\ar[dd, "i_*"] & \\
    &  & H_{\lambda}(V\union D_L, V \union \qty{D_L\setminus \theset{q} }) \ar[dd, "h_2"] \\
    H_\lambda(W)\ar[rrddd]  & \\
    &  &  H_\lambda(V\union D_L, V) \ar[dd] \\
    \\
    &  & H_\lambda(W, V) \\
    \end{tikzcd}
    \end{center}

    We have $H_\lambda(V') \ni h_0([M]) = (S_R \cdot M) \cdot \Phi(\alpha)$ where $\alpha \in H_0(S_R)$ is the canonical generators and

    \begin{align*}
    \Phi: H_0(S_R) \to H_\lambda (V, V'\setminus S_R)
    \end{align*}

    is the Thom isomorphism.

    Thus $h_3 h_2 h_1(\Phi(\alpha)) = D_L(p)$.