Let $f \colon X \to Y$ and $g \colon Y \to Z$ be maps of spaces.

If $f$ and $g$ are $n$-connected, then so is $g f$.

If $f$ is $(n-1)$-connected and $g f$ is $n$-connected, then $g$ is $n$-connected.

If $g$ is $(n+1)$-connected and $g f$ is $n$-connected, then $f$ is $n$-connected.

Proposition

Let

$\begin{matrix}
B & \longleftarrow & A & \longrightarrow & C \\
g \downarrow & & \downarrow f & & \, \downarrow h \\
B' & \longleftarrow & A' & \longrightarrow & C'
\end{matrix}$

be a commutative diagram of maps of spaces. If $f$ is $(n-1)$-connected and $g$ and $h$ are $n$-connected, then the induced map between homotopy pushouts$B \sqcup_A^h C \to B' \sqcup_{A'}^h C'$ is $n$-connected.

$\begin{matrix}
Y & \longrightarrow & X & \longleftarrow & Z \\
g \downarrow & & \downarrow f & & \, \downarrow h \\
Y' & \longrightarrow & X' & \longleftarrow & Z'
\end{matrix}$

be a commutative diagram of maps of spaces. If $f$ is $(n+1)$-connected and $g$ and $h$ are $n$-connected, then the induced map between homotopy pullbacks$Y \times_X^h Z \to Y' \times_{X'}^h Z'$ is $n$-connected.

References

Tammo tom Dieck, Algebraic topology. European Mathematical Society, Zürich, 2008.

Revised on September 7, 2012 02:34:43
by Toby Bartels
(98.23.143.147)