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(A) Show that a space \(X\) is contractible if and only if every map \(f:X \to Y\), for arbitrary \(Y\), is nullhomotopic. (B) Similarly, show \(X\) is contractible if and only if every map \(f:Y \to X\) is nullhomotopic.
(A) Assume a space \(X\) is contractible. Then there exists a singleton \(\{
p \} = : P\) and maps \(\alpha:X \to \{ p \}\) and \(\beta:\{ p \} \to X\)
such that \(\alpha\beta \simeq \mathbbm{1}_{P}\) and \(\lambda :=
\beta\alpha \simeq \mathbbm{1}_{X}\). Let \(F:X \times I \to X\) denote
the homotopy with \(F(x,0) = x\) and \(F(x,1) = \lambda(x)\).
Let \(Y\) be an arbitrary space with a map \(f:X \to Y\). Then \(H:=f \circ
F\) is a homotopy with \(H(x,0) = f(x)\) and
\(H(x,1) = f(\lambda(x))\) which is constant \(\forall x \in X\).
Hence, \(f\) is nullhomotopic.
Conversely, assume that every map \(f:X \to Y\) is nullhomotopic for any
arbitrary \(Y\). Then there is some \(x_{0} \in X\) and \(\phi:X \to X\) such
that \(\phi(X) = \{ x_{0} \}\). Then \(\phi:X \to \{ x_{0} \}\) and
\(\iota_{X}:\{ x_{0} \} \to X\) form a homotopy equivalence as
\(\phi\iota_{X} = \mathbbm{1}_{\{ x_{0} \}}\) and \(\phi\iota_{X} = \phi
\simeq \mathbbm{1}_{X}\). Hence, \(X\) is contractible.
(B) Assume a space \(X\) is contractible. Define \(F\) and \(\lambda\) as above in (A). Let \(Z\) be an arbitrary space with a map \(g:Z \to X\). Then \(G:=F \circ (g \times \mathbbm{1}_{I})\) is a homotopy (since \(g \times \mathbbm{1}_{I}\) is continuous on \(Z \times I\)). Further, \(\forall x \in X\), \(G(x,0) = F(g(x),0) = g(x)\) and \(G(x,1) = F(g(x),1) = \lambda(g(x))\). Hence, \(g\) is nullhomotopic. Conversely, assume that every map \(g:Z \to X\) is nullhomotopic for any arbitrary \(Z\). Then \(\mathbbm{1}_{X}\) is nullhomotopic and you can apply the same argument (A) to show \(X\) is contractible.
(B) Assume a space \(X\) is contractible. Define \(F\) and \(\lambda\) as above in (A). Let \(Z\) be an arbitrary space with a map \(g:Z \to X\). Then \(G:=F \circ (g \times \mathbbm{1}_{I})\) is a homotopy (since \(g \times \mathbbm{1}_{I}\) is continuous on \(Z \times I\)). Further, \(\forall x \in X\), \(G(x,0) = F(g(x),0) = g(x)\) and \(G(x,1) = F(g(x),1) = \lambda(g(x))\). Hence, \(g\) is nullhomotopic. Conversely, assume that every map \(g:Z \to X\) is nullhomotopic for any arbitrary \(Z\). Then \(\mathbbm{1}_{X}\) is nullhomotopic and you can apply the same argument (A) to show \(X\) is contractible.