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\begin{center}
  {\large Mathematics 101}\\Fall 2014\\Homework 3
\end{center}

\begin{enumerate}
\item (Pullbacks) Given a ring $R$ with identity and $R$-modules $ A,
  B, C$, consider the following diagram with $R$-linear maps $\alpha,
  \beta$:
  \[
  \xymatrix{& A \ar[d]^\alpha \\
    B \ar[r]_\beta & C }
  \]
  A \textit{pullback} for this diagram (also called a fiber product of
  $\alpha$ and $\beta$) consists of the following data:

  \begin{compactitem}
  \item An $R$-module $P$ and $R$-linear maps $\alpha': P \to A$,
    $\beta': P \to B$ making the following diagram commute ($\alpha
    \circ \alpha' = \beta\circ \beta'$).
    \[
    \xymatrix{
      P \ar[d]_{\beta'} \ar[r]^{\alpha'} & A \ar[d]^\alpha \\
      B \ar[r]_{\beta} & C }
    \]
  \item It satisfies the following universal mapping property: For
    every $R$-module $X$ admitting a commutative diagram of linear
    maps ($\alpha\circ \varphi = \beta\circ \psi$), there is a unique
    $R$-linear map $h : X \to P$ such that the following diagram
    commutes (that is, $\psi = \beta' \circ h$ and $\varphi = \alpha'
    \circ h$):
    \[
    \xymatrix{ X \ar@/_/ @{->} [ddr]_{\psi} \ar@{-->}[dr]^h
      \ar@/^/[drr]^{\varphi} \\
      &P \ar[d]_{\beta'} \ar[r]^{\alpha'} & A \ar[d]^\alpha \\
      & B \ar[r]_\beta & C }
    \]

    \vfill

    Pullbacks appear frequently in topology where one constructs a
    pullback of a fiber bundle.  Now to the exercise: Let $P = \{ (a,
    b) \in A \times B \mid \alpha(a) = \beta(b)\}$, $\alpha'$ and
    $\beta'$ the standard projections to the factors $A$ and $B$.
    Show that $P$ together with the associated data form a pullback,
    i.e., verify that $P$ is an $R$-module and that the above
    universal mapping property holds for this choice of $P$ and maps
    $\alpha', \beta'$.  \smallskip

    Remark: Dummit and Foote (exercise 27, 10.5) introduce the
    pullback as a subset of the direct sum $A\oplus B$. Categorically,
    you should find the direct product a more useful perspective. And
    of course note that the pullback is unique up to isomorphism by
    virtue of its definition by universal mapping property.
  \end{compactitem}

  \newpage

\item Verify the following are examples of pullbacks.
  \begin{enumerate}
  \item With $\alpha'$ and $\beta'$ the obvious projections:
    \[
    \xymatrix{
      A\times B \ar[d]_{\beta'} \ar[r]^-{\alpha'} & A \ar[d]^\alpha \\
      B \ar[r]_{\beta} & 0 }
    \]

  \item For $\varphi: B \to C$ an $R$-linear map,
    \[
    \xymatrix{
      Ker(\varphi) \ar[d]_{\subseteq} \ar[r] & 0 \ar[d] \\
      B \ar[r]_{\varphi} & C }
    \]
  \item For $\varphi: B \to C$ an $R$-linear map and $A$ a submodule
    of $C$,
    \[
    \xymatrix{
      \varphi^{-1}(A) \ar[d]_{\subseteq} \ar[r] & A \ar[d]^\subseteq \\
      B \ar[r]_{\varphi} & C }
    \]

  \end{enumerate}

\item Let
  \[
  \xymatrix{
    P \ar[d]_{\beta'} \ar[r]^{\alpha'} & A \ar[d]^\alpha \\
    B \ar[r]_{\beta} & C } \mbox{ be a pullback of }
  \xymatrix{& A \ar[d]^\alpha \\
    B \ar[r]_\beta & C }
  \]
  \begin{enumerate}
  \item Show that if $\alpha$ is injective, so is $\beta'$.
  \item Show that if $\alpha$ is surjective, so is $\beta'$.
  \end{enumerate}

\item In the commutative diagram,
  \[
  \xymatrix{
    Q\ar[d]_{\alpha''}\ar[r]^{\gamma'}&P \ar[d]_{\alpha'} \ar[r]^{\beta'} & A \ar[d]^\alpha \\
    C\ar[r]_{\gamma}&B \ar[r]_{\beta} & D }
  \]
  Show that if $\alpha\circ \beta' = \beta\circ\alpha'$ and
  $\alpha'\circ\gamma' = \gamma\circ \alpha''$ are pullbacks, show
  that $\alpha\circ (\beta'\circ\gamma') = (\beta\circ
  \gamma)\circ\alpha''$ is a pullback.

\item (Schanuel's lemma) Let $R$ be a ring, and consider two exact
  sequences of $R$-modules
  \[\xymatrix{%
    0 \ar@{>}[r]& K \ar@{>}[r] &P \ar@{>}[r]^\varphi&M\ar@{>}[r] &0& 0
    \ar@{>}[r]& K' \ar@{>}[r] &P' \ar@{>}[r]^{\varphi'}&M\ar@{>}[r] &0
  }
  \]
  where $P$ and $P'$ are projective.  Show that as $R$-modules $K'
  \oplus P \cong K \oplus P'$.  \newline

  \textit{Hint:} Show there is an exact sequence
  \[\xymatrix{%
    0 \ar@{>}[r]& \textrm{ker}\; \pi \ar@{>}[r] &X
    \ar@{>}[r]^\pi&P\ar@{>}[r] &0 }
  \]
  with $\textrm{ker}\; \pi \cong K'$ and where $X$ is the fiber
  product of $\varphi$ and $\varphi'$ as in the first problem.  From
  this deduce that $X \cong K' \oplus P$.  Similarly, show $X \cong K
  \oplus P'$.

\item (Pushouts) Given a ring $R$ with identity and $R$-modules $ A,
  B, C$, consider the following diagram with $R$-linear maps $\alpha,
  \beta$:
  \[
  \xymatrix{ C \ar[r]^\alpha \ar[d]_\beta & A\\
    B \\}
  \]
  A \textit{pushout} for this diagram consists of the following data:

  \begin{compactitem}
  \item An $R$-module $P$ and $R$-linear maps $\alpha': A \to P$,
    $\beta': B \to P$ making the following diagram commute ($\alpha'
    \circ \alpha = \beta'\circ \beta$).
    \[
    \xymatrix{
      C \ar[d]_{\beta} \ar[r]^{\alpha} & A \ar[d]^{\alpha'} \\
      B \ar[r]_{\beta'} & P }
    \]
  \item It satisfies the following universal mapping property: For
    every $R$-module $X$ admitting a commutative diagram of linear
    maps ($\varphi\circ \alpha = \psi\circ \beta$), there is a unique
    $R$-linear map $h : P \to X$ such that the following diagram
    commutes (that is, $\psi = h \circ\beta' $ and $\varphi = h\circ
    \alpha'$):

    \[
    \xymatrix{
      C \ar[d]_{\beta} \ar[r]^{\alpha} & A \ar[d]^{\alpha'}  \ar@/^/[ddr]^{\varphi}\\
      B \ar@/_/ @{->} [drr]_{\psi}\ar[r]_{\beta'} & P  \ar@{-->}[dr]^h\\
      &&X}
    \]


    Define the pushout as follows: Using the universal mapping
    property of the coproduct, each pair of homomorphisms $\varphi: A
    \to X$ and $\psi: B \to X$ can be described by a single
    homomorphism $\omega: A\oplus B \to X$.  Let $\iota_A : A \to
    A\oplus B$ and $\iota_B: B \to A\oplus B$ be the standard
    injections.  Put $P = (A\oplus B)/K$ where $K =
    Im(\iota_A\circ\alpha - \iota_B\circ \beta) = \{ (\alpha(c),
    -\beta(c)) \in A\oplus B \mid c\in C\}$ and let $\pi: A\oplus B
    \to P$ be the quotient map.  Put $\alpha' = \pi\circ \iota_A$ and
    $\beta' = \pi \circ \iota_B$.  Show that $\alpha'\circ \alpha =
    \beta'\circ \beta$ is a pushout.


    \[
    \xymatrix{
      C \ar[d]_{\beta} \ar[r]^{\alpha} & A \ar[d]^{\iota_A} \\
      B \ar[r]_{\iota_B} & A\oplus B  \ar@{->}[dr]^\omega\ar[r]^\pi&P\ar@{-->}[d]^h\\
      &&X}
    \]

    Remark: You will use the notion of a pushout in the proof of the
    Siefert-van Kampen theorem.
  \end{compactitem}

\end{enumerate}




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