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\pagestyle{fancy} \lhead{{\sf Homework 11 $|$ Due November 14 (Thursday)
}} \rhead{\thepage} \cfoot{{\sf MthSc 4560/6560 $|$ Topology $|$ Fall
    2013 $|$ M.~Macauley}}



\begin{document}

$\;$

\noindent Throughout, $X$ is a path-connected topological space. Given
a path $\alpha\colon I\to X$ with endpoints $\alpha(0)=x_0$ and
$\alpha(1)=x_1$, define the map
\[
\hat\alpha\colon\pi_1(X,x_0)\longto\pi_1(X,x_1)\,,\qquad
\hat\alpha([f\,])=[\bar\alpha]*[f\,]*[\alpha]\,.
\]
If $h\colon(X,x_0)\longto(Y,y_0)$ is continuous, then the
\emph{induced homomorphism} relative to $x_0$ is
\[
h_*\colon\pi_1(X,x_0)\longto\pi_1(Y,y_0)\,,\qquad h_*([f\,])=[h\circ f\,]\,.
\] 

\medskip

\begin{enumerate}

% m52.2
\item Let $\alpha\colon I\to X$ be a path from $x_0$ to $x_1$, and
  $\beta\colon I\to X$ a path from $x_1$ to $x_2$. Show that if
  $\gamma=\alpha*\beta$, then
  $\hat\gamma=\hat\beta\circ\hat\alpha$. \\



% m52.3
\item Show that $\pi_1(X,x_0)$ is abelian if and only if for every
  pair $\alpha$ and $\beta$ of paths from $x_0$ to $x_1$, we have
  $\hat\alpha=\hat\beta$. \\


% Mendelson Exercise 4.7.4, Hatcher 0.9
\item Recall that a \emph{retraction} of $X$ onto a subset $A$ is a
  continuous map $r\colon X\to A$ such that $r\circ r=r$. Equivalently,
  this means that $ri=1_A$, where $i\colon A\into X$ is the inclusion
  map and $1_A$ is the identity mapping on $A$.
  
  \begin{enumerate}[(a)]
  \item Let $a_0\in A$. Prove that if $r\colon X\to A$ is a
    retraction, then $i_*\colon\pi_1(A,a_0)\to\pi_1(X,a_0)$ is
    one-to-one and $r_*\colon\pi_1(X,a_0)\to\pi_1(A,a_0)$ is onto.
  \item Prove that a retract of a contractible space is contractible. \\
  \end{enumerate}


% m52.5
\item Let $A$ be a subspace of $\R^n$ and $h\colon(A,a_0)\longto(Y,y_0)$
  be continuous. Show that if $h$ extends to a continuous map of
  $\R^n$ into $Y$ (that is, if there is some $f\colon\R^n\to Y$ such
  that $f\circ i=h$ for the inclusion map $i\colon A\to\R^n$), then
  $h_*$ is the trivial homomorphism. \\


% m52.5
  \item Show that the induced homomorphism of a continuous map is
    independent of base point, up to isomorphism. More precisely, let
    $h\colon X\to Y$ be continuous, with $h(x_0)=y_0$ and
    $h(x_1)=y_1$. Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$,
    and let $\beta=h\circ\alpha$. Show that 
    \[
    \hat\beta\circ(h_{x_0})_*=(h_{x_1})_*\circ\hat\alpha\,.
    \]
    In other words, the following diagram commutes:
    \[
    \xymatrix{\pi_1(X,x_0)\ar[rr]^{(h_{x_0})_*}\ar[d]_{\hat\alpha} &&
      \pi_1(Y,y_0)\ar[d]^{\hat\beta}
      \\ \pi_1(X,x_1)\ar[rr]^{(h_{x_1})_*} && \pi_1(Y,y_1)}
    \]

  


\end{enumerate}

\end{document}
