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\begin{document}
\parindent 0in


\parbox[b]{2in}{\tiny
  \href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}: \href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
  \newline \href{http://www.math.toronto.edu/~drorbn/classes/\#2425}{2024-25}: \href{http://drorbn.net/25-1301}{MAT 1301 Algebraic Topology}:
}~{\LARGE\bf Homework Assignment 6}
\hfill\parbox[b]{2in}{\tiny
  \null\hfill\sheeturl
  \newline\null\hfill Due March 12 at 11:59pm
}~%
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  \qrcode[height=1.2em,level=L,nolink]{drorbn.net/25-1301}
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\begin{center}\begin{minipage}{6.5in}


\noindent{\bf Problem 1.} Just in case you thought commutative diagrams cannot get any worse, here's a question to prove you wrong (though actually, it is surprisingly easy).
\begin{enumerate}
\item Define a category $\calS$ of ``short exact sequences of chain complexes''.
\item Define a category $\calL$ of ``long exact sequences''.
\item Construct a functor $\calH\colon\calS\to\calL$. (We've constructed this functor on objects already, so you don't need to do that again. The challenge is to do it on morphisms and to verify that starting from a morphism $\calF$ in $\calS$, its image $\calH(\calF)$ is indeed a morphism in $\calL$).
\end{enumerate}

\vskip 2mm
\noindent{\bf Problem 2.} Given a morphism $f\colon(X,A)\to(Y,B)$, show that the diagram
\[ \xymatrix{
  H_n(X,A) \ar[r]^\delta \ar[d]_{f_\ast} &amp; H_{n-1}(A) \ar[d]_{f_\ast} \\
  H_n(Y,B) \ar[r]^\delta &amp; H_{n-1}(B)
} \]
is commutative.

\vskip 2mm
\noindent{\bf Problem 3.} Given a triple of spaces $B\subset A\subset X$, construct a long exact sequence relating $H_\ast(X,A)$, $H_\ast(A,B)$, and $H_\ast(X,B)$.

\vskip 2mm
\noindent{\bf Problem 4.} Khovanov homology is an invariant of knots obtained by first defining a chain complex $\calC$, and then taking its homology. To show that Khovanov homology does not change when the knot moves, one has to show that pieces of $\calC$ can be cancelled off --- be removed without changing the homology.
\begin{enumerate}
\item Suppose $\calC'$ is an acyclic subcomplex of $\calC$ (``acyclic'' is a different word for ``exact'', or ``having no homology''). Show that $\calC'$ can be cancelled off. Namely, that the quotient $\calC/\calC'$ has the same homology as the original complex $\calC$.
\item Suppose $\calC'$ is a subcomplex of $\calC$, and that the quotient complex $\calC/\calC'$ is acyclic. Then $\calC'$ has the same homology as $\calC$ (roughly, the complement of $\calC'$ can be cancelled off).
\item There is a third theorem along these lines. What does it say?
\end{enumerate}

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