I'm having troubles with this problem:

Let $A$ be a closed subset of $X$ and $B$ a subset of $Y$. Assume that $(X,A)$ has the homotopy extension property with respect to $B$ and that $(X\times I,X\times \partial I\cup A\times I)$ has the homotopy extension property with respect to $Y$. Prove that if $f:(X,A)\rightarrow (Y,B)$ is homotopic (as a map of pairs) to a map which sends all of $X$ to $B$, then it is homotopic relative to $A$ to such a map."

Now, my problem is that I can't see how to use the (HEP) with respect to $B$ and $Y$ to prove this homotopy must be relative to $A$.

I'm not looking for solution, hints or any kind of help I will appreciate it.

The reference used is Edwin H. Spanier Algebraic topology.

  • $\begingroup$ Where precisely does Spanier state this? $\endgroup$
    – Paul Frost
    Oct 9, 2018 at 17:24
  • $\begingroup$ P57 problem D.4 $\endgroup$ Oct 9, 2018 at 19:06

1 Answer 1


We have a homotopy of pairs $H : (X,A) \times I = (X \times I, A \times I) \to (Y,B)$ such that $H_0 = f$ and $H_1(X) \subset B$. Here, $H_t(x) = H(x,t)$.

We shall find a homotopy of pairs $$G : ((X,A) \times I) \times I = (X \times I \times I, A \times I \times I) \to (Y,B)$$ such that

(1) $G_0 = H$

(2) $(G_s)_0 = f$ for all $s \in I$

(3) $G(X \times \{ 1 \} \times I) \subset B$

(4) $H' = G_1$ is a homotopy such that $H'_t \mid_A = f \mid_A$ for all $t \in I$.

Here, $G_s : (X,A) \times I \to (Y,B)$ is the homotopy defined by $G_s(x,t) = G(x,t,s)$.

Then obviously $H'$ is the desired homotopy.

By the HEP with respect to $Y$, it suffices to find $$\phi : (X \times \partial I \cup A \times I) \times I \to Y$$ such that

(1) $\phi(x,t,0) = H(x,t)$ for all $(x,t) \in X \times \partial I \cup A \times I$

(2) $\phi(x,0,s) = f(x)$ for all $x \in X, s \in I$

(3) $\phi(X \times \{ 1 \} \times I \cup A \times I \times I) \subset B$

(4) $\phi(a,t,1) = f(a)$ for all $(a,t) \in A \times I$.

To get $\phi$, you have to proceed in three steps:

(a) Find a homotopy $\phi^A : A \times I \times I \to B$ such that $\phi^A(a,t,0) = H(a,t)$, $\phi^A(a,t,1) = f(a)$ for all $(a,t) \in A \times I$ and $\phi^A(a,0,s) = f(a)$ for all $a \in A$, $s \in I$.

(b) Use the HEP with respect to $B$ to find a homotopy $\phi^1 : X \times I \to B$ such that $\phi^1(x,0) = H_1(x)$ for all $x \in X$ and $\phi^1(a,s) = \phi^A(a,1,s)$ for all $a \in A$, $s \in I$.

(c) Use $\phi^A$ and $\phi^1$ to define $\phi$ and check continuity.


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