Vtyjkyuk

We form $X$ by attaching a $3$-cell to $S^2$ by a degree $2$ map, and form $Y$ be attaching a $3$-cell to $S^2$ by a degree $4$ map. I am trying to find out whether the identity $S^2 to S^2$ extends to a map $Y to X$. I have already showed that it doesn't extend to a map $X to Y$ because the induced map $mathbbZ cong H_2(S^2) to H_2(Y) cong mathbbZ/4$ sends $2$ to $2$, but if it factored as $H_2(S^2) to H_2(X) to H_2(Y)$, it would have to send $2$ to $0$ because $H_2(X) cong mathbbZ/2$. No such obstruction exists when considering $Y to X$.

I feel such a map couldn't exist since you can't "unravel" $Y$ to only a double twist. But, I can't find an obstruction. Any help would be appreciated. Thanks.

For each integer $k$ write $k:S^2rightarrow S^2$ for the degree $k$ self-map. Then, since $4simeq 2circ 2$ we have a homotopy commutative diagram

$requireAMScd$
beginCD
S^2@>4>> S^2@>p>>Y\
@V2 V V @VVid_S^2 V@VV theta V\
S^2 @>2>> S^2@>q>>X
endCD

where each row is the cofiber sequence that defines the spaces $X$, $Y$, and $theta:Yrightarrow X$ is the induced map of cofibers. The map $theta$ is clearly what you are looking for.

Let's do this in a bit more detail. First observe that $X$, $Y$ are the topological pushouts in the diagrams
$requireAMScd$
beginCD
S^2@>4>> S^2\
@Vi V V @VVpV\
D^3 @>l>> Y
endCD$requireAMScd$
beginCD
S^2@>2>> S^2\
@Vj V V @VVqV\
D^3 @>k>> X
endCD

where $i=j$ are the boundary inclusions (they are the same map, but we'll assign them different symbols to make clear to which we are referring).

First we note that since the composite $jcirc2:S^2rightarrow S^2rightarrow D^3$ is null-homotopic and $i$ is a cofibration, there exists and extension $underline 2:D^3rightarrow D^3$ such that $underline 2circ i=jcirc 2$. This then gives us a map $kcircunderline 2:D^3rightarrow X$ that satisfies

$$(kcircunderline 2)circ i=kcirc(underline 2circ i)=kcirc (jcirc 2)=(kcirc j)circ 2=(qcirc 2)circ 2=qcirc (2circ 2).$$

Now choose a homotopy $F:2circ 2simeq 4$ so that $qcirc (2circ2)simeq qcirc 4$. Using again the fact that $i$ is a cofibration we get a homotopy $G:D^3times Irightarrow X$ starting at $G_0=kcirc underline 2$ and satisfying $G_tcirc i= qcirc F_t$ for each $tin I$. In particular its end map $G_1:D^3rightarrow X$ now satisfies

$$G_1circ i= qcirc F_1=qcirc 4.$$

It follows using the that $Y$ is the pushout of the maps $i$ and $4$ that there is an induced map $theta:Yrightarrow X$ satisfying

$$thetacirc p=qcirc id_S^2,qquad thetacirc l=G_1simeq kcirc underline 2$$

and the first of these properties is exactly what you are looking for.

It may help you to visualise things if you put the pushout square for $Y$ above the pushout square for $X$ and draw in all the maps we've used.

Finally note that although the identity map of $S^2$ does not extent to a map $Xrightarrow Y$, the degree $2$ map does. You may like to follow the steps we gone through here to write down the map it induces.