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I am trying to solve an exercise in Hatcher "Algebraic Topology" but am a little confused by the terminology he is using (just so you know, it is exercise 5 in chapter 1.1). He writes that in the problem 'homotopic' means 'homotopic without regard to basepoints.' In general, Hatcher defines a homotopy between two paths $f, g:I\to X$ with the same endpoints as a continuous function $H:I\times I \to X$ such that $H(-, 0) = f$ and $H(-, 1)=g$ and that $H(0,t)=f(0)$ for all $t$ and $H(1, t) = f(1)$ for all $t$. So when Hatcher says 'homotopic without regard to basepoints' does he mean we can relax the requirement that $f$ and $g$ have the same endpoints in the definition of 'homotopic'?

Thanks for the help!

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He defines homotopy with respect to basespoints as a function $H$ which satisfies the conditions you mention and such that $H(0,t)=H(1,t)=\text{basepoint}$ for all $t$. The version without regards to basepoints is then relaxing this last condition and only demanding that $H(0,t)=H(1,t)$ for all $t$ (so that all intermediate paths are actually closed paths) – Mariano Suárez-Alvarez May 17 '12 at 6:59
Oh yes of course, sorry I forgot to add that requirement, I have edited my question to accommodate your remark. – student555 May 17 '12 at 7:03

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