Is every map between connected CW-complexes homotopic to a map sending $x$ to $y$?

Question

Let $f \colon X \to Y$ be a continuous map between connected CW-complexes $X$ and $Y$.

Choose points $x \in X$ and $y \in Y$. Is $f$ homotopic to a map sending $x$ to $y$?

Answer 1

In short: Suppose $f: X\to Y$ is any map with $f(x) = y^{\prime}$ and let $\gamma:[0,1]\to Y$ be a path between $\gamma(0)=y^{\prime}$ and $\gamma(1)=y$. Then $f$ and $\gamma$ together constitute a continuous map $$X\times [0,1]\supset X\times\{0\}\cup_{\{x\}\times \{0\}} \{x\}\times [0,1]\to Y.$$ Since the inclusion $\{x\}\hookrightarrow X$ is a cofibration, this map extends to a map $X\times [0,1]\to Y$, which can be interpreted as a homotopy from $f$ to a map $X\to Y$ sending $x$ to $y$.

More details: First, without loss of generality, we can assume that $x$ is a $0$-cell of the given CW-structure by potentially passing to a subdivision. It is therefore sufficient to show that the any inclusion $A\hookrightarrow X$ of a subcomplex $A$ of a CW-complex $X$ is a cofibration. The corresponding extension problem $$(\ddagger)\qquad X\times [0,1]\supset X\times\{0\}\cup_{A\times \{0\}} A\times [0,1]\to Y$$ can then inductively be reduced to the special case $X={\mathbb D}^n\supset A={\mathbb S}^{n-1}$, where it holds since the inclusion on the LHS of $(\ddagger)$ is a deformation retract.

Reference: Hatcher, Algebraic Topology, Proposition 0.16

Answer 2

Yes, this is true and to prove this, just follow the proof of the Cellular approximation theorem.