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The question is, does the proof from Hatcher given below still work if we take out the part about $f^{-1}(x)$ being compact?

$\hskip 0.5in$ Hatcher

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up vote 5 down vote accepted

In the proof it is shown that, given a loop $f$ in $\mathbb{S}^n$ with basepoint $x_0$ and another point $x\neq x_0$, it's possible to produce a loop $g$ homotopic to $f$ which avoids $x$. It's completely possible that the set $f^{-1}(B)$ consists of infinitely many disjoint open intervals $(a_i,b_i)$, but the compactness of $f^{-1}(x)$ guarantees that $f$ hits $x$ only on finitely many of those intervals. Hence we need to deform $f$ only on finitely many of the intervals $(a_i,b_i)$.

Suppose we did not know that $f^{-1}(x)$ is compact and accordingly try to deform $f$ on every interval $(a_i,b_i)$. More precisely, given a homotopy $H_i:[a_i,b_i]\times\mathbb{I}\to\mathbb{S}^n$ deforming $f_i$ to $g_i$ (where $f_i$ is the restriction of $f$ to $[a_i,b_i]$ and $g_i$ is a path inside $B$ from $f(a_i)$ to $f(b_i)$ which avoids $x$) for each $i$, we define $$H:\mathbb{I}\times\mathbb{I}\to\mathbb{S}^n,H(s,t)= \begin{cases}H_i(s,t), & s\in[a_i,b_i] \\ f(s), & \text{otherwise}. \end{cases}$$ Note that this is a completely well-defined function, but the problem is that there is no guarantee that $H$ is continuous, since possibly we have manipulated $f$ on infinitely many intervals $[a_i,b_i]$.

The compactness of $f^{-1}(x)$ solves this problem, because if we deform $f$ only on finitely many intervals, then we can apply the following general proposition to demonstrate the continuity of $H$. If $X=F_1\cup F_2\ldots\cup F_k$ where $F_i$'s are closed and $f$ is such a function that the restrictions of $f$ to each $F_i$ are continuous, then $f$ is continuous.

Added: Why manipulating $f$ on infinitely many intervals may cause problems with respect to continuity? I hope that the following example is not too irrelevant to the question at hand. At least it shows that if we change a continuous function on infinitely many intervals in such a way that all the restrictions of the new function to these intervals remain continuous, then in general we cannot conclude that the new function is continuous.

Let $f$ be the zero function on $[0,1]$ and consider the countably infinite collection of intervals $(\frac{1}{n+1},\frac{1}{n}),n\in\mathbb{N}$. On each interval change $f$ so that the graph of the new function, say $g$, on this interval will be a $\wedge$-shaped bump with maximum $1$ and infimum $0$. In particular, $g$ restricted to any of the intervals will be continuous. However, $g$ is not continuous at $0$, since the midpoints of the intervals are taken to $1$ but $g(0)=0\neq 1$.

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Sorry, I should have been more specific with my question. Why does manipulating $f$ on more than finitely many intervals cause a problem re. continuity? – Dilitante Nov 12 '11 at 12:24
Nice explanation though – Dilitante Nov 12 '11 at 12:26

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