Solution Verification of Exercise 1.1.2 Algebraic Topology Hatcher

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Exercise 1.1.2

Show that the change-of-basepoint homomorphism $\beta_h$ depends only on the homotopy class of $h$.

Solution

I have to show that, if $h,g: I \to X$, where $h(0) = x_0 = g(0)$ and $h(1) = x_1 = g(1)$, are path such that $h \simeq g$ then $\beta_h = \beta_g$.

In fact, observe that, proof that $\beta_h = \beta_g$ is equivalent to show that $\beta_h^{-1} \circ \beta_g = id$. Furthmore, $\beta_h^{-1} = \beta_{\tilde{h}}$.

Note that, if $h \simeq g$ then $\tilde{g} \cdot h \simeq e_{x_1}$, because $h \simeq g \Rightarrow \tilde{g} h \simeq \tilde{g} g \simeq e_{x_1}$. Mutatis Mutandis, we can show that $h \simeq g \Rightarrow \tilde{h} g \simeq e_{x_1}$.

So, back to the original problem, we have
$$
\beta_h^{-1} \circ \beta_g \left[ f\right] = \beta_h^{-1} ( \beta_g \left[ f\right]) = \beta_{\tilde{h}} ( \beta_{g} \left[ f\right]) = \beta_{\tilde{h}} \left[g \cdot f \cdot {\tilde{g}} \right] = \left[{\tilde{h}} \cdot g \cdot f \cdot {\tilde{g}} \cdot h \right] = \left[{\tilde{h}} \cdot g \right] \left[f\right] \left[ {\tilde{g}} \cdot h \right] = \left[e_{x_1} \right] \left[f\right] \left[e_{x_1}\right] = \left[f \right]
$$

Best Answer

I think your solution is valid.

At this point in the book, you already know that a change of basepoint homomorphism $\beta_h: \pi_1(X, x_0) \to \pi_1(X, x_1)$ is an isomorphism, and you already know that the inverse of $\beta_h$ is $\beta_{\widetilde{h}}$ (where $\widetilde{h}$ is the reverse of the path $h$). So proving that $\beta_g = \beta_h$ is equivalent to proving that $\beta_{\widetilde{h}} \circ \beta_{g} = \text{id}$.

Personally I would have used the notation $\bar{h}$ to denote the reverse of $h$, since the notation $\widetilde{h}$ is suggestive of a lift to a covering space. But that's very minor.

Also, at the very end, you wrote $ \left[{\tilde{h}} \cdot g \cdot f \cdot {\tilde{g}} \cdot h \right] = \left[{\tilde{h}} \cdot g \right] \left[f\right] \left[ {\tilde{g}} \cdot h \right] = \left[e_{x_1} \right] \left[f\right] \left[e_{x_1}\right] = \left[f \right] $. This calculation is legitimate, but I'm not sure why you decided it was necessary to split $\left[{\tilde{h}} \cdot g \cdot f \cdot {\tilde{g}} \cdot h \right] $ into $ \left[{\tilde{h}} \cdot g \right] \left[f\right] \left[ {\tilde{g}} \cdot h \right]$. It would have been perfectly fine to write $ \left[{\tilde{h}} \cdot g \cdot f \cdot {\tilde{g}} \cdot h \right] = \left[e_{x_1} \cdot f \cdot e_{x_1}\right] = \left[f \right].$ Indeed, what I just wrote is the true crux of the calculation.

Finally, while your proof strategy is valid, I would suggest that it would have been simpler to prove $\beta_g = \beta_h$ directly. Indeed, if $f$ is an arbitrary element of $\pi_1(X, x_0)$, then $$ \beta_g(f) = [g . f . \widetilde{g}] = [h . f . \widetilde{h}] = \beta_h(f).$$

If you have doubts about these calculations (either mine or your own), then perhaps it would be useful to clearly articulate the "magic fact" that is underpinning these calculations:

Suppose that $f$ and $g$ are paths from $x_1$ to $x_2$, and suppose there exists a homotopy of paths from $f$ to $g$. (Note that by definition, a homotopy of paths is a continuous deformation that leaves the endpoints fixed.) Then:

  • If $h$ is a path from $x_0$ to $x_1$, then there exists a homotopy of paths from $h . f$ to $h . g$
  • If $l$ is a path from $x_2$ to $x_3$, then there exists a homotopy of paths from $f . l$ to $g . l$.

Often, the phrase "there exists a homotopy of paths between $f$ and $g$" is abbreviated to $f \simeq g$. And if $f$ and $g$ are loops, then that phrase might also be abbreviated to $[f] = [g]$. But fundamentally, all we're doing in our calculations is using known homotopies of paths to infer the existence of other homotopies of paths, by applying the "magic fact" over and over again.

My calculation applies the "magic fact" twice:

  • $g \simeq h$, so $\widetilde{g} \cdot f \cdot g \simeq \widetilde{g} \cdot f \cdot h$.
  • $\widetilde{g} \simeq \widetilde{h}$, so $\widetilde{g} \cdot f \cdot h \simeq\widetilde{h} \cdot f \cdot h$.

If I counted correctly, your calculation applies the "magic fact" six times. You also call upon two other well-known facts: the fact that $\widetilde{g} . g \simeq \widetilde h . h \simeq e_{x_1}$, and the fact that $e_{x_1} . l \simeq l . e_{x_1} \simeq l$ for any loop $l$ at $x_1$. So every step is legitimate. There's nothing mysterious going on.

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