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Let $f,g:S^1\rightarrow X$ be paths with $g$ constant. At the exercise 3.3 of Rotman's "Introduction to Algebraic Topology" I need to show that $$ f\simeq g \ \ \mathrm{rel}\{1\} \Leftrightarrow f\simeq g, $$ where $1$ is the $\ e^{2\pi i 0}\in S^1 \ $ and $\ "\simeq"\ $ is the homotopic relation. The $\ "\Rightarrow"\ $ direction seems to me trivial, the $"\mathrm{rel}\{1\}"$ element adds that, if $F$ is the homotopy function, then $F(1,t)=f(1)=g(1)$, but the $\ "\Leftarrow " \ $ got me baffled. From the supposition $f$ and $g$ being free homotopic, ultimately, as long as I get it right, the closed loop $f$ has to deform continuously into one of its point that coicides with the constant.

Can you provide some further insight, or some indications in order to help me make progress?

Edit: In this particular exercise, $f(1)=g(1)$ is given.

knstans
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    Try to show that for any based loops $l,g$, if $l$ is homotopic to $g$ then $l$ is homotopic to $g$ conjugated by the loop that the basepoint takes in the homotopy from $l$ to $g$. – Connor Malin Nov 23 '19 at 02:20

1 Answers1

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Suppose that $f \simeq g$, so there exists a continuous map $H : S^1 \times [0,1] \to X$ such that $H(z,0)=f(z)$ and $H(z,1)=x_0$.

Constructing a homotopy $f \simeq q$ rel $\{1\}$ will require using two quotient maps.

Consider first the quotient map $q : S^1 \times [0,1] \to D^2$ defined by $$q(x,t) = (1-t) x $$ This quotient map takes $S^1 \times 1$ to the origin and is otherwise one-to-one. Since $H$ is constant on $S^1 \times 1$, it follows that $H$ factors through this quotient map: $$H : S^1 \times [0,1] \xrightarrow{q} D^2 \xrightarrow{h} X $$ with $H = h \circ q$. Also, the restriction $h \mid S^1$ is our original map $f$.

Now we need another quotient map, with the same domain and range $p : S^1 \times [0,1] \to D^2$, but with a different effect:

  1. $p(x,0)=x$ for all $x \in S^1$
  2. $p(x,1)=p(1,t)=1$ for all $x \in S^1$, $t \in [0,1]$
  3. Except for the identifications in 2, $p$ is one-to-one.

Here's a formula for $p$: $$p(x,t) = (1-t) x + t $$

The composition $S^1 \times [0,1] \xrightarrow{p} D^2 \xrightarrow{h} X$ is a homotopy which shows that $f \simeq g$ rel $\{1\}$.

Lee Mosher
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