Show that if $X$ is path connected then every path $f:I\to X$ is homotopic to a constant path $g(t)=x$. Does this mean show that there is an $x$ (or even for all $x$) such that this is true for all paths $f$, or to show for each $f$, there is an $x$ such that this is true? I know that two paths are homotopic if there exists a continuous map $H(s,t):I\times I\to X$ with $H(s,0)=f(s)$ and $H(s,1)=g(s)$. I think I'm just misunderstanding the question. If we take an annulus, which is path connected, and consider a loop around the inner circle, then how is this loop, which is a path, homotopic to any constant loop? Could somebody explain what the question is asking since I can't make any progress with it until I understand. I thought googling this question would tell me something and nothing came up. Is the claim false or am I misunderstanding something?
Show that if $X$ is path connected then every path $f:I\to X$ is homotopic to a constant path $g(t)=x$
algebraic-topologygeneral-topology
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Best Answer
This can work (to assuage your doubts) as the homotopy is a "free" one, where the homotopy does not have to preserve any fixed point, as in the fundamental group, which is why the torus or circle don't work as counterexamples..
If $f$ is a path, we can define $H: [0,1] \times [0,1] \to X$ by $H(s,t) = f(s(1-t))$ which is clearly continuous, and for $t=0$ we just have the path $f$ and for $t=1$ a constant map with constant value $f(0)$. We could have an arbitary $x$ for the constant because in path connected space $X$ any two constant maps are homotopic (and homotopy is an equivalence relation).