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\lhead{{\sf Summer Bridge Course (Algebra)}: Thursday July 15}
\chead{}\rhead{\thepage}
\lfoot{Due Friday, July 17 2026}
\cfoot{}\rfoot{Written by M.~Macauley}
\pagestyle{fancy}


\begin{document}

%\(\;\) \vspace{-4mm}


\noindent \textbf{Topics}: Universal properties \\

\noindent \textbf{Do}: Answer the following questions. \\

%%--------------------------------------------------------------------- 

  %% HW 8, Problem 1
\begin{exercise}
  Suppose we have functions \(g_i\colon X\to Y\) and \(h_i\colon Y\to Z\) between sets, for \(i=1,2\).
  \begin{enumerate}[label=(\roman*)]
  \item Show that if \(g\) is surjective, it right-cancels:
    \[
    h_1\circ g=h_2\circ g\quad\Longrightarrow\quad h_1=h_2.
    \]
  \item Show that if \(h\) is injective, it left-cancels: \quad
    \[
    h\circ g_1=h\circ g_2\quad\Longrightarrow\quad g_1=g_2.
    \]
  \end{enumerate}
\end{exercise}

%%--------------------------------------------------------------------- 

  %% HW 8, Problem 2
\begin{exercise}
  Let \(X\) and \(U\) be vector spaces, \(Y\leq X\) a subspace with \(\pi\colon X\to X/Y\) the canonical quotient map.
  \begin{enumerate}[label=(\roman*)]
  \item Prove the universal property of quotient maps for vector spaces: for any linear map \(f\colon X\to U\) satisfying \(\Ker(\pi)\leq\Ker(f)\), there exists a unique linear map \(h\colon X/Y\to U\) such that \(f=h\circ\pi\), as shown in the following commutative diagram:
    \[
    \begin{tikzpicture}[ scale=3.5, >=latex,
        every path/.style={line width=.6pt},
        every label/.style={font=\footnotesize}
      ]
      \begin{scope}[shift={(0,0)}]
        \node (X) at (0,0) {\(X\)};
        \node (U) at (1,0) {\(U\)};
        \node (X/Y) at (.5,-.5) {\(X/Y\)};
        \draw[->] (X) -- (U) node[midway,above]{\(f\)};
        \draw[->] (X) -- (X/Y) node[midway,below left]{\(\pi\)};
        \draw[->,dashed] (X/Y) -- (U) node[midway,below right]{\(h\)};
      \end{scope}
      %%
      \begin{scope}[shift={(2,0)}]
        \node (X) at (0,0) {\(x\)};
        \node (U) at (1,0) {\(f(x)\)};
        \node (X/Y) at (.5,-.5) {\(\overline{x}\)};
        \draw[->] (X) -- (U) node[midway,above]{\(f\)};
        \draw[->] (X) -- (X/Y) node[midway,below left]{\(\pi\)};
        \draw[->,dashed] (X/Y) -- (U) node[midway,below right]{\(h\)};
      \end{scope}
    \end{tikzpicture}
    \]
  \item Prove that $h$ is injective if and only if \(\Ker(\pi)=Y\).
  \item Use the previous parts to prove that first isomorphism theorem for vector spaces: if $f\colon X\to U$ is a linear map, then
    \[
    X/\Ker(f)\cong\Image(f),\qquad\text{where}\quad \Image(f)=f(X).
    \]
  \end{enumerate}
\end{exercise}

%%--------------------------------------------------------------------- 

%% HW 8, Problem 3
\begin{exercise}
  Prove the universal property of subspace inclusions for vector spaces: for any linear map \(f\colon U\to X\) satisfying \(\Image(f)\leq Y\), there exists a unique linear map \(h\colon U\to Y\) such that \(f=\iota\circ h\), as shown in the
  following commutative diagram:
  \[
  \begin{tikzpicture}[ scale=3.5, >=latex,
      every path/.style={line width=.6pt},
      every label/.style={font=\footnotesize}
    ]
    \begin{scope}[shift={(0,0)}]
      \node (U) at (0,0) {\(U\)};
      \node (X) at (1,0) {\(X\)};
      \node (Y) at (.5,-.5) {\(Y\)};
      \draw[->] (U) -- (X) node[midway,above]{\(f\)};
      \draw[->,dashed] (U) -- (Y)
      node[midway,below left]{\(h\)};
      \draw[->] (Y) -- (X)
      node[midway,below right]{\(\iota\)};
    \end{scope}
    %%
    \begin{scope}[shift={(2,0)}]
      \node (U) at (0,0) {\(u\)};
      \node (X) at (1,0) {\(f(u)\)};
      \node (Y) at (.5,-.5) {\(f(u)\)};
      \draw[->] (U) -- (X) node[midway,above]{\(f\)};
      \draw[->,dashed] (U) -- (Y)
      node[midway,below left]{\(h\)};
      \draw[->] (Y) -- (X)
      node[midway,below right]{\(\iota\)};
    \end{scope}
  \end{tikzpicture}
  \]  
\end{exercise}

%%---------------------------------------------------------------------  

\end{document}