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This exercise is from Kosniowski's Algebraic Topology, page 137, exercise 15.11, d.

Show that two continuous maps $\varphi, \psi :X \rightarrow Y$ with $\varphi(x_0) = \psi (x_0)$ for some $x_0 \in X$ induce the same homomorphism from $\pi(X,x_0)$ to $\pi(Y, \varphi(x_0))$ if $\varphi$ and $\psi$ are homotopic relative to $x_0$.

We need to show that, given the equivalence class of a loop with base $x_0$, $[g]\in \pi(X,x_0)$, then $[\varphi \circ g ] = [\psi \circ g ]$.

The fact that $\varphi(x_0) = \psi (x_0)$ means that $[\varphi \circ g]$ and $[\psi \circ g]$ are both equivalence classes of loops in $Y$ with base point $\varphi(x_0)$. But I don't see how the fact that $\varphi$ is homotopic to $\psi$ implies that $[\varphi \circ g]=[\psi \circ g]$.

Yagger
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1 Answers1

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You can prove this more general result:

Lemma. Let $A\subset X,$ $B\subset Y.$ If $f_0,f_1:X\to Y$ and $g_0,g_1:Y\to Z$ are maps, $f_0\simeq f_1$ rel $A,$ $g_0\simeq g_1$ rel $B,$ and $f_0(A)=f_1(A)\subset B,$ then $g_0\circ f_0\simeq g_1\circ f_1$ rel $A.$

Proof sketch. If $h:X\times I\to Y$ is a homotopy from $f_0$ to $f_1$ rel $A$ and $h':Y\times I\to Z$ is a homotopy from $g_0$ to $g_1$ rel $B,$ check that $H:X\times I\to Z$ given by $$H(x,t)=h'(h(x,t),t)$$ gives the desired homotopy from $g_0\circ f_0$ to $g_1\circ f_1$ rel $A.$

positron0802
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