Suppose we are given a pullback diagram of topological spaces
$\require{AMScd}$
\begin{CD}
A @>>> B\\
@VVV @VVV\\
C @>>>D
\end{CD}
and we want to look at the fibres of the maps $A\to C$ and $B\to D$. Then I know that the induced maps in the fibres are an isomorphism for each basepoint.
However, I fail to see why this is true.
Can someone elaborate on this??
Best Answer
The fiber of $A \rightarrow C$ at a point $c$ is given by the pullback $$\begin{array}{ccc} F & \rightarrow & A\\ \downarrow && \downarrow\\ * & \rightarrow & C \end{array}$$
By pasting this square to the left of your square and using that pullbacks compose we find that the composite square is (isomorphic to) the fiber of $B \rightarrow D$ at $f(c)$.
Similarly, for a point $d$ in the image of $f$, we can compute the fiber by considering the pullback square $$\begin{array}{ccccc} E & \rightarrow & \rightarrow & \rightarrow & B\\ \downarrow &&&& \downarrow\\ * &\rightarrow & C & \rightarrow & D \end{array}$$ By using the universal property of your pullback square, the upper map factors via $A$ and we get two squares pasted together, the right one and composite one being pullbacks. Hence the left square is a pullback by the cancellation property, showing that the fiber of $B \rightarrow D$ at $d$ is isomorphic to the fiber of $A \rightarrow C$ for any $c$ with $f(c)=d$.