For sets $X$ and $Y$, let $X\to Y$ denote the set of functions from $X$ to $Y$. It is clear that, for a topological space $X$ there is a bijection between $([0,1]\times [0,1]) \to X$ and $[0,1] \to ([0,1] \to X)$ given by associating to $H: [0,1]\times [0,1] \to X$ the function $s\mapsto (t\mapsto H(s,t))$. In this assignment, we shall see that, if we use the appropriate topology on paths in a space $X$, i.e., maps from $[0,1]$ to $X$, this is in fact a bijection between homotopies and paths of paths.

Given topological spaces $X$ and $Y$, the compact open topology on the set $C(X, Y)$ of continuous functions from $X$ to $Y$ is the topology with sub-base constructed as follows. For $K\subset X$ compact and $U\subset Y$ open, we define $V(K, U)$ to be the set of continuous functions $f: X\to Y$ with $f(K)\subset U$. The sets $V(K, U)$ form a sub-base for the compact-open topology.

Assume henceforth that, for spaces $X$ and $Y$, $C(X, Y)$ has the compact-open topology.

1. Assume that $X$ is a compact topological space and $Y$ is a metric space. Let $f_n: X\to Y$ be a sequence of maps and $f: X\to Y$ be a map. Show that the sequence $f_n$ converges to $f$ in the compact open topology if and only if it converges uniformly to $f$.

2. Let $X$ be a topological space. Show that a function $H: [0,1]\times [0,1] \to X$ is continuous if and only if

   • for each $s\in [0,1]$, $f_s: t\mapsto H(s, t)$ is continuous.
   • the function $s\mapsto f_s$ is a continuous function from $[0,1]$ to $C([0,1], X)$ with the compact open topology.

As mentioned earlier, in algebraic topology we associated to spaces algebraic objects. The first half of this course will be focused on one such - the fundamental group.

We shall first construct the so called fundamental groupoid, so that lemmas can be proved in the generality we need. A groupoid is like a group, except that not all pairs of elements can be multiplied. A typical example is the set of invertible linear maps between subspaces of a vector space $V$, with composition as the multiplication - this is defined in the case of the codomain of the second map coinciding with the domain of the first. There are identities corresponding to each subspace, and each element is invertible in an appropriate sense.

The construction of the fundamental groupoid is in two steps. We first define a multiplication on the space of paths of a space. We then define an equivalence relation on this space, so that the multiplication makes the quotient into a groupoid.

Paths and Multiplication

Fix a topological space $X$. For points $a,b\in X$, let $\Omega(X; a, b)$ be the set of paths from $a$ to $b$, i.e., $\Omega(X; a,b) = \{\alpha: [0,1] \to X : \text{$\alpha$ continuous}, \alpha(0)=a, \alpha(1) = b\}.$

For $a, b, c\in X$, we have a multiplication from $\Omega(a, b)\times \Omega(b,c)\to \Omega(a,c)$ which corresponds to a path from $a$ to $b$ followed by a path from $b$ to $c$ (precise formulas will be given in the lectures).

Let $\Omega(X)$ be the union of the sets $\Omega(X; a,b)$. We then get a partially defined product $*$ on this from the above products. Note that this is not associative and there are no identity elements or inverses.

Homotopy between paths and the Groupoid structure

A crucial idea is to regard paths as equivalent if one can be deformed to the other fixing end points. This gives an equivalence relation, where the deformation is called a path homotopy, or simply a homotopy. Formally a path homotopy is a map $H : [0,1] \times [0,1] \to X$. For paths $\alpha$ and $\beta$, we denote path homotopy by $\alpha\sim \beta$.

Let $\pi(X)$ denote the quotient of $\Omega(X)$ by the equivalence relation given by path homotopy, with $\pi(X, a,b)$ as obvious. Then the product on $\Omega(X)$ induces one on $\pi(X)$. Moreover, we see that this product is associative.

Furthermore, if $e_a$ denotes the constant path at $a$, then if $\alpha$ is a path from $a$ to $b$, we have $e_a * \alpha \sim \alpha$ and $\alpha * e_b \sim \alpha$. Finally, if $\bar\alpha : t \mapsto \alpha(1-t)$, then $\alpha * \bar\alpha \sim e_a$ and $\bar\alpha * \alpha \sim e_a$. This means that $*$ makes $\pi(X)$ into a groupoid.

Fundamental group

Now pick a basepoint $x\in X$. We define the fundamental group of $X$ with basepoint $x$ to be $\pi_1(X, x)= \pi(X; x, x)$. It is straightforward to see that the groupoid structure on $\pi(X)$ makes $\pi_1(X, x)$ into a group.

Change of basepoint

Suppose we choose a different basepoint $x’$ for the space $X$, then if there is a path $\alpha$ from $x$ to $x’$, we see that the $\pi_1(X, x)$ and $\pi_1(X, x’)$ are isomorphic. Namely, we use the groupoid multiplication to construct a function $ \beta\mapsto\alpha * \beta * \bar\alpha$ from $\pi_1(X, x’)$ to $\pi_1(X, x)$. Replacing $\alpha$ by $\bar\alpha$ gives a function in the opposite direction. Finally the groupoid properties show that these are inverses of each other.

We begin with a typical topology problem (or rather, collection of problems).

Extension problems:

Assume we are given topological spaces $X$ and $Y$, a subspace $A\subset X$ and a continuous function $f: A\to Y$. Is there an extension of $f$ to $F: X\to Y$, i.e., if $i:A\to X$ is the inclusion function, is there a function $F: X\to Y$ so that $f = F\circ i$?

Two specific instances of this:

1. $X = [-1,1]$, $A = \{-1,1\}$, $Y =\mathbb{R}$, $f: A\to Y$ is given by $f(x)= x$, for $x=-1,1$.

2. $X = [-1,1]$, $A = \{-1,1\}$, $Y =\mathbb{R} \setminus {0}$, $f: A\to Y$ is given by $f(x)= x$, for $x=-1,1$.

It is not hard to see that the first problem has a positive solution and the second a negative one - in the first case we use a convex function, while the second follows using the fact that the image of a connected set is connected. We shall return to general methods after considering some more problems.

We next describe some features of the extension problem that holds for many topological problems (we use a term from logic for this that can be safely ignored).

First order topological problems:

• These are formulated in terms of topological spaces and continuous functions (which we call maps) : both the data and unknowns are in terms of these.
• The typical conditions are equalities between various compositions of functions.
• Conditions may also be in terms of topological properties of the spaces.

More topological problems

Homeomorphisms problem:

Given spaces $X$ and $Y$, are there continuous functions $f: X\to Y$ and $g: Y \to X$ with compositions the identities on $X$ and $Y$?

Lifting problem

Given spaces $D$, $X$ and $Y$ and maps $p: Y\to X$ and $f: D\to X$, is there a function $F: D \to Y$ with $f = p\circ F$?

Solving topological problems

As we can already see in the above examples of extension problems, showing existence and non-existence involve very different methods. Existence is typical based on constructions, generally using geometry and point-set topology.

On the other hand, the above non-existence result could be done using point-set topology because of the special nature of the spaces - with one of them disconnected. To generalize this, we view connectivity as the $0$-dimensional case of algebraic topology.

What is algebraic topology?

The idea of algebraic topology is to map a (first order) topological problem to an algebraic one, with spaces mapped to groups (or other algebraic objects) and continuous functions mapped to homomorphisms. We can generally conclude that if a topological existence problem has a solution, then so does the corresponding algebraic problem.

Often simple algebraic arguments show that there is no solution to the algebraic, hence the topological, problem. For instance, if two spaces are homeomorphic, then their associated groups are isomorphic - and hence if we obtain different groups we can conclude that the given spaces are not homeomorphic.