What is wrong with this proof that the identity map of $S^1$ is nullhomotopic?
Matthew Harrington 
I have read that the identity map of the unit circle $S^1$ is not nullhomotopic. In fact, I am very new to the subject, so I wonder what is wrong with the following reasoning (that seems to suggest the opposite):
If the identity map $i : S^1 \to S^1$ is nullhomotopic, there exists a homotopy $F : S^1 \times I \to S^1$ with $F(x,0)=x$ and $F(x,1)=k$, for all $x \in S^1$. But this, in other words, just means that $x$ and $k$ can be joined by a path, for any $x \in S^1$ - which is true, since $S^1$ is path connected.
After all, can't we think of a homotopy of two functions $f, g : X \to Y$ as a function that for all $x \in X$, gives us a path between $f(x)$ and $g(y)$?
I wonder where is the contradiction? What I am missing?
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$\begingroup$It is true that if you have a homotopy $F : X \times [0,1] \to Y$ between $f,g : X \to Y$, then for all $x \in X$ this defines a path between $f(x)$ and $g(x)$, given by $t \mapsto f(x,t)$.
This is a "if... then..." statement. The converse need to be true. If I give you a map $F : X \times [0,1] \to Y$, and I tell you that for all $x$, $t \mapsto F(x,t)$ is continuous; in other words, I give you a collection of paths from $f(x)$ to $g(x)$ for all $x$; does it imply that $F$ itself is continuous? Of course not, and it's rather easy to construct counterexamples.
There is an intuitive way to see this for the circle. Imagine your circle is a clock. Let's say you want to find a homotopy between the identity and the map that's constantly equal to 12 o'clock (i.e. $i \in S^1 = \{ z \in \mathbb{C} \mid |z| = 1 \}$). Then for each point $x$ of the circle you have to a path from $x$ to 12:00. Either you can go through the left part of the circle (going clockwise), or you can go through the right part (going counterclockwise).
It's rather intuitive that if you have a point that just a little bit before 12:00, you will want to go clockwise. And if you're a little bit after 12:00, you will want to go counterclockwise. Since you want this whole thing to be continuous, this will extend all the way to the bottom. But what about 6 o'clock now? How do you choose clockwise or counterclockwise? Whichever way you choose, you will "rip" the circle in half, because just a little bit to the left or to the right you will be going in the other direction. And intuitively, ripping things in half isn't a continuous thing.
This isn't a formal proof, obviously. There is a lot of theory to actually prove that it's impossible to find a homotopy between the identity and a constant map, e.g. by showing that the fundamental group of the circle is $\mathbb{Z}$, which isn't trivial. But this is the intuition: if you have to choose continuously paths from each $x \in S^1$ to a fixed $x_0 \in S^1$, you will have to "rip" the circle in half sooner or later, and this isn't continuous.
$\endgroup$ $\begingroup$Let us recall the formal definition of a homotopy between $f$ and $g$. Suppose $f,g : X\to Y$ are continuous functions, then a homotopy $H: X \times I \to Y$ is a continuous function such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for all $x\in X$. What's the difference between this and what you've mentioned? Your condition isn't enough to satisfy that we get a continuous map. It is true that one can fix a point and vary $t$ to get a path.
So why is the identity map not null-homotopic? One can show that if the identity map on a space is nullhomotopic then the space is contractible, $S^1$ is not contractible, so here's your contradiction. I'm not sure of an easy way to see it without this lemma. One could consider that if $f: S^1 \to S^1$ is nullhomotopic then there exists $\bar{f}: D^2 \to S^1$ which restricted to $S^1$ is $f$, but this gives a retraction of $D^2$ onto its boundary. This uses the fundamental groups of $S^1$ and $D^2$ though.
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