Cohomology- Homotopy

Imagine you are stuck in a strange planet without light source and it is always dark. You landed in an area with some rocks and some pleasant flat ground (otherwise, how else could you have landed?). Your job is to direct an incoming ship to a flat ground.

Since you took some algebraic topology at college, you are a bonafide Mathematician and would always think as one. So using your knowledge, you design this highly contrived, non-optimal "rock" detection system. (If you need more convincing Mathematician's way of doing certain simple things is unique, please refer to Mathematician's way of making Tea- Check out fantastic online book http://www.topologywithouttears.net).

Your plan or device has the following steps.
1. Put a peg on the ground at a random place. Tie a rope.
2. Walk in some direction for some time unrolling your rope, assuming that you haven't hit a rock, plant another peg where you tie other end of the rope.
3. Pause and recognize this rope as a curve on the surface and name it $f$.
4. Walk a few feet and repeat the same steps - if you are successful unrolling your rope, you have another curve - call is $g$.

Now if you can drag rope $f$ to rope $g$, you don't have a rock in between otherwise you have an obstruction or a rock!

If you can drag $f$ to $g$ (or vice versa), you call it $f ~ g$ and this basic idea of homotopy.

Now while dragging the rope from $f$ to $g$, you are working along a map $F$ which at start (we call the start as $t=0$ as a parameter for the drag), the map should be $f$ and at the end (say $t=1$) $F$ should be $g$.

This notion is formalized as follows. Let $M,N$ be two manifolds and let $f,g:M\rightarrow N$ be two smooth functions. If there is a $C^\infty $ map
\begin{equation}
  F: M \times R \rightarrow N
\end{equation}
such that $F(M,0) = f$ and $F(M,1)=g$, then we say $f$ is homotopic to $g$ and write $f \tilde g$.

Now that we have such a map, we can add extra nomenclature when certain conditions occur. Let $N=\mathcal{R}^n$. $F:M\times \mathcal{R} \rightarrow \mathcal{R}^n$ linear in $t$ can be expressed as
\begin{equation}
  F(M,t) = f(1-t)+(t)g
\end{equation}
When $t=0$ implies $F(M,0)=f$ and when $t=1$, $F(M,1)=g$.

Notice $F(M,t)=f+(g-f)t$ which is like $y=mx+c$ or a straight line equation in terms of $t$. Such a straight line path is called "Straight line homotopy".

Convex means that any arbitarary two points can be joined by a straight line.  On any subset of $\mathcal{R}^n$ for which this is true, straight line homotopy is applicable.

Sometimes we are fortune enough to get maps $f,g$ where $g:N \rightarrow M$ such that $f\circ g$ is identity on $N$ and $g \circ f$ is identity on $M$. In such situations, $M$ is considered to be "homotopy equivalent" to $N$. $M,N$ are said to be same homotopy type.

Notice, nothing precludes $N$ to be a single point set. For example, if $M=\mathcal{R}^n$ and $N$ is a single point set, we can smoothly scrunch $\mathcal{R}^n$ to a single point. Such manifolds that can be shrunk to a point are called "contractible".