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How do I show that if $X$ is simply connected, then there is a unique path class in $X$ with initial point $x$ and terminal point $y$?

Doesn't this just follow from the definition of simply connectedness?

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How do you define simply connected? – mixedmath Nov 8 '11 at 2:31
It depends on what you mean by "just". It certainly follows from that definition, else you wouldn't be able to show it; but how simple would you require the proof to be to call it "just"? – joriki Nov 8 '11 at 2:31
mixedmath: A space is simply connected if every loop (basepoint arbitrary) is null-homotopic. @joriki Sorry I should have written "immediately" instead of "just". – stefan Nov 8 '11 at 2:44
@stefan: That leaves the same question -- how immediate is "immediately"? One person's "immediate" is another person's several steps. – joriki Nov 8 '11 at 2:50
up vote 4 down vote accepted

It is not completely immediate. You want to show that if you have two paths $\gamma_1$ and $\gamma_2$ both from $x$ to $y$, then there's a homotopy from $\gamma_1$ to $\gamma_2$.

You can construct a closed loop as $\gamma_1-\gamma_2$ and (since the space is simply connected) let that contract to a point -- but in the intermediate stages of that contraction you have no guarantee that $x$ and $y$ will still be hit (indeed it cannot be the case that both are hit all the way, unless $x=y$). So you need to do some plumbing and splicing in order to derive a homotopy between $\gamma_1$ and $\gamma_2$ from the null-homotopy. For complete rigor, showing the homotopy as an explicit function defined by a case split is probably in order.

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