Thread: Relative Homotopy of paths

Just a little bit confused over the notion of homotopy of paths relative to a subset, let's say I have two paths f and g from [0,1] to a space X, and they're homotopic relative to a subspace A of X. If I were able to graph these out, am I correct in saying that they can meander away from each other as they go, but as they pass through A, they are essentially the same line, i.e. one is laid over the other?

And then for the question of fundamental groups, if I have two loops in the same equivalence class, are they basically just one laid on top of the other? Or can there be a situation where two loops take different paths but are in the same class?

Thanks

A homotopy of paths relative to a subspace A of X is a homotopy which fixes the points of A, so yes what you're saying is correct but unorthodox/unrigorous. And by 'graphing these out', I hope you mean graphing the homotopy at each point $\displaystyle t\in [0,1]$, because this infinite class of functions must fix A as well.

For the question on fundamental groups, the latter is true. If two loops are the same equivalence class, you can write a homotopy from one loop to another (and the converse is also true). Intuitively, you can continuously deform from one to another without passing through 'holes'. For example, since the complex plane is contractible (so has trivial fundamental group), every conceivable loop you can think of basepointed at the same point, say the origin, are in the same equivalence class. On the other hand in $\displaystyle \mathbb{C}-\{0\}$, the loop $\displaystyle e^{it}$ basepointed at 1 and the constant loop $\displaystyle 1$ are not homotopic (in different equivalence classes) since the hole at the origin prevents the first loop from shrinking to a point.

Ah great thanks a million, I understand it now, was just getting a bit confused with notations etc. Just one last question you could possibly be able to help me with? I'm studying from my professor's notes, and at one point he proves the following proposition:

Let $\displaystyle p:{\~X} \rightarrow X$ be a covering map, let Z be a connected topological space, and let $\displaystyle g:Z \rightarrow {\~X}$ and $\displaystyle h: Z \rightarrow {\~X}$ be continuous maps. Suppose that $\displaystyle p \circ g = p \circ h$ and that $\displaystyle g(z) = h(z)$ for some $\displaystyle z \in Z$. Then $\displaystyle g = h$.

I understand the proof of this, but as he moves on to results about paths in covering spaces, he continually references the fact, following from the above proposition, that the lift to $\displaystyle {\~X}$ of any path in $\displaystyle X$ is uniquely determined by its starting point.

I'm just a little bit unsure as to how this result follows from the proposition?

Thanks again!

When you take a lift, you send the basepoint $\displaystyle x_0$ to one of its pre-images in $\displaystyle p^{-1}(x_0)$. Clearly the lift depends on which one you send it to.

Now take $\displaystyle Z$ to be $\displaystyle I=[0,1]$ and note that paths in $\displaystyle \tilde{X}$ are defined as functions $\displaystyle I\to \tilde{X}$.