The induced homomorphisms of maps homotopic relative to a subset are the same

Question

A theorem from my book states that:

Let $\phi_0, \phi_1 : X \to Y$ be maps that are homotopic relative to the subset $\{x\}$. Then $${\phi_0}_* = {\phi_1}_* : \pi (X,x) \to \pi (Y, \phi_0(x))$$ i.e. the induced homomorphisms are the same

Apparently the proof is immediate; but I don't see how I would go about proving this. Intuitively; I can start to see why this is true, as the two maps being homotopic means that they can be 'continuously deformed' into each other, so the induced homomorphisms should be the same, but I'm not sure how to actually prove this fact.

Answer 1

Recall that for a map $\phi:X\to Y$, the induced map $\phi_*:\pi_1(X,x_0)\to\pi_1(Y,\phi(x_0))$ is given by $$\phi_*[\omega] = [\phi\circ\omega].$$ We have two maps $\phi_1,\phi_2:(X,x_0)\to (Y,y_0)$, and a homotopy $H:X\times I\to Y$ satisfying $$H(x,0)= \phi_1(x),\quad H(x,1)=\phi_2(x),\quad H(x_0,t) = y_0.$$ What we want to show is that $$[\phi_1\circ\omega] = \phi_{1*}[\omega] = \phi_{2*}[\omega] = [\phi_2\circ\omega],\quad \forall[\omega]\in\pi_1(X,x_0).$$ That is, given any loop $\omega:I\to X$ based at $x_0$, $\phi_1\circ\omega\simeq \phi_2\circ\omega$. All we have to do is form the composition $$H':\quad I\times I\xrightarrow{\ \omega\times id_I \ }X\times I\xrightarrow{\ H \ }Y$$ which gives a homotopy between $\phi_1\circ\omega$ and $\phi_2\circ\omega$ (check this!).