Math and stuff

Proposition Show that the isomorphism $\pi_1(X \times Y) \approx \pi_1(X) \times \pi_1(Y)$ in this proof is given by \([f] \mapsto (p_{1*}([f]), p_{2*}([f]))\) where $p_1$ and $p_2$ are the projections of $X \times Y$ onto its two factors. Solution In the proof, we defined $\phi: \pi_1(X \times Y, (x_0, y_0)) \rightarrow...

Proposition Given a map $f: X \rightarrow Y$ and a path $h: I \rightarrow X$ from $x_0$ to $x_1$, show that \(f_*\beta_h = \beta_{fh}f_*\). Solution Let $[g] \in \pi_1(X, x_1)$ be given. Then $g$ is a loop in $X$ based at $x_1$. \[\begin{align*} f_*\beta_h([g]) &= f_*([h \cdot g \cdot \overline{h}])...

Proposition Given a space $X$ and a path-connected subspace $A$ containing the basepoint $x_0$, show that the map $\pi_1(A, x_0) \rightarrow \pi_1(X, x_0)$ induced by the inclusion $A \rightarrow X$ is surjective iff every path in $X$ with endpoints in $A$ is homotopic to a path in $A$. Solution First,...

Proposition Show that the change-of-basepoint homomorphism $\beta_h$ depends only on the homotopy class of $h$. Solution Let $h \simeq g$. By the inverse lemma, $\overline{h} \simeq \overline{g}$. The product operation respects homotopy classes, $h \cdot f \cdot \overline{h} \simeq g \cdot f \cdot \overline{g}$. Thus $[h \cdot f \cdot \overline{h}]...

Problem Statement Show that composition of paths satisfies the following cancellation property: If $f_0 \cdot g_0 \simeq f_1 \cdot g_1$ and $g_0 \simeq g_1$ then $f_0 \simeq f_1$. Solution By the inverse lemma, $\overline{g_0} \simeq \overline{g_1}$. We know that $f_0 \cdot g_0 \simeq f_1 \cdot g_1$. As mentioned on P.26,...