Homotopy classes and Induced map being injective I've been solving the exercise 3 of Chapter 15 Section 2 in Topology written by James Dugundji.
The exercise question is:

Let $\phi:X \rightarrow Y$ be continuous. Prove $\phi^{*} :[Y,Z] \rightarrow [X,Z]$ is injective for every Z iff there is a continuous $\rho:Y \rightarrow X$ such that $\phi \circ \rho \simeq 1_Y.$

The $\phi^{*}$ is defined as $\phi^{*}[f] = [f \circ \phi]$ where $[f]$ denotes the homotopy class of $f$.
I proved a similar exercise but with surjectivity and the $\Leftarrow$ by using that $f \circ \phi \simeq g \circ \phi$ whenever $f \simeq g$. However, with the $\Rightarrow$, my friends and I have been stuck for several hours not even coming up with any results. Seems like this is supposed to be a routine but we found this quite challenging.
So my question is:

*

*Is $(\Rightarrow)$ right?

*Any hints or counterexamples regarding this exercise?

 A: The statement is quite false as it stands.
For my example I'll need some facts not contained in Dugundji's book. Unfortunately, I know of no simpler construction. Hatcher's book and Arkowitz's book are good references for what follows.

Fact: If $\phi:X\rightarrow Y$ is a basepoint-preserving map between connected CW complexes and for some path-connected pointed space $Z$ the induced map $\phi^*:[Y,Z]_\ast\rightarrow[X,Z]_\ast$ of pointed homotopy sets is injective, then the induced map $\phi^*:[Y,Z]\rightarrow[X,Z]$ of unpointed homotopy sets is also injective.

Note that the converse fails. In any case, if $X$ is a connected CW complex and $Z$ any space, then two maps $f,g:X\rightarrow Z$ can be homotopic only if $f(X)$ and $g(X)$ are contained in the same path-component $Z_0$ of $Z$. In this case, $f\simeq g$ if and only if the two restrictions $f',g':X\rightarrow Z_0$ are homotopic.
Thus if we take $X=S^n\times S^n$ and $Y=S^{2n}$, and let
$$\phi:S^n\times S^n\rightarrow S^{2n}$$
be the map pinching to the top cell, then in the sequel it will be sufficient to work with path-connected pointed spaces and assume all maps and homotopies preserve basepoints. I'll continue to use $[-,-]_*$ to denote pointed homotopy classes of pointed maps.
Now, we have a cofiber sequence.
$$S^{2n-1}\xrightarrow{\omega} S^n\vee S^n\xrightarrow{i}S^n\times S^n\xrightarrow\phi S^{2n}$$
where $i$ is the inclusion of the $(2n-1)$-skeleton and $\omega$ is the Whithead product. The sequence extends to the right in the standard manner, with the next map being the suspension $\Sigma\omega:S^{2n}\rightarrow\Sigma(S^n\vee S^n)\cong S^{n+1}\vee S^{n+1}$. However,
$$\Sigma (S^n\times S^n)\simeq S^{n+1}\vee S^{n+1}\vee S^{2n+1},$$
and the map $\Sigma i:\Sigma(S^n\vee S^n)\rightarrow \Sigma(S^n\times S^n)$ has a right homotopy inverse $\alpha$. Consequently
$$\Sigma\omega\simeq(\alpha\circ\Sigma i)\circ\omega\simeq\alpha\circ(\Sigma i\circ\Sigma\omega)\simeq\alpha\circ\Sigma(i\circ\omega)\simeq\alpha\circ\ast\simeq\ast$$
since $i,\omega$ are adjancent maps in a cofiber sequence.
On the other hand, the segment
$$S^n\times S^n\xrightarrow\phi S^{2n}\xrightarrow{\Sigma\omega}S^{n+1}\vee S^{n+1}$$
is a cofiber sequence, so if $Z$ is any space and $f:S^{2n}\rightarrow Z$ a map with $\phi^*(f)=f\circ\phi\simeq\ast$, then there is $g:S^{n+1}\vee S^{n+1}\rightarrow Z$ such that $f\simeq g\circ\Sigma\omega$. Of course $\Sigma\omega$ being trivial forces $f\simeq\ast$.

Proposition: $\phi^*:[S^{2n},Z]_*\rightarrow[S^n\times S^n,Z]_*$ is injective for any space $Z$. $\quad\blacksquare$

On the other hand, the map $\phi:S^n\times S^n\rightarrow S^{2n}$ induces the trivial map on all homotopy groups. Thus if $\rho:S^{2n}\rightarrow S^n\times S^n$ is any map, then $\phi\circ\rho\simeq\ast$.

Proposition: $\phi$ has no right homotopy inverse. $\quad\blacksquare$

