I am currently reading the proof of proposition 1B.9 of Hatcher's "Algebraic Topology" on page 90. On page 91 he states:
If two maps $f, f': (X, x_0) \to (Y, y_0)$ induce the same homomorphism $f_\ast, f_\ast'$ on $\pi_1$, then we see immediately that their restrictions to $X^1$ are homotopic, fixing $x_0$.
$X^1$ is the 1-skelton of the connected cw complex $X$ and we assume that $X$ has a single 0-cell. I want to give more details to the proof and started like this: Let $\gamma_e: I \to X^1$ be a loop around $x_0$ walking on the closure of a 1-Zell $e \in X^1$. Then we have $f_\ast([\gamma_e]) = [f \circ \gamma_e] = [f' \circ \gamma_e] = f_\ast'([\gamma_e])$ and $f$ and $f'$ are homotopic on the closure of each 1-cell hence on all of $X^1$.
Now to my question: Hatcher writes that the resulting map is $H: X^1 \times I \cup X \times \partial I \to Y$. I don't understand how he gets this domain. I think $X^1 \times I$ makes sense as $H$ is a homotopy between $f|_{X^1}, f'|_{X^1}$ but why the union with $X \times \partial I \to Y$? I can't imaging $H$ being continuous that way going from all of $f, f'$ to their restrictions.