UPDATE: The original question can be simplified to this: Given a finite simplicial complex $X$, can I find a continuous $f : X \rightarrow \text{skel}^2(X)$ that fixes $\text{skel}^1(X)$? Basically, can I smoosh $X$ down to its $2$-skeleton without touching its $1$-skeleton?
This seems to be possible for simple examples. For instance, consider a simplicial solid sphere. Starting at the boundary, we can "push" the outer $2$-face of a $3$-cell against its other three faces, resulting in three $2$-cells, and we can repeat this until there are no more $3$-cells. Notice that each such operation does not move the $1$-skeleton, and the result still lives inside the $2$-skeleton. As long as there is a $3$-cell with a face that does not belong to any other $3$-cell, this approach should work with any $3$-dimensional simplicial complex.
There are cases where there are no $3$-cells that don't share a face with another $3$-cell, such as a simplicial $3$-sphere. In this case, we can take a $3$-cell and map it onto the rest of the complex, resulting in a simplicial $3$-disk, after which we can repeatedly apply the trick above to obtain the desired $f$. However, I'm not sure what to do in the case of the $3$-torus ($S^1 \times S^1 \times S^1$). I cannot do the same thing I did with the $3$-sphere, since the $3$-torus does not retract to the punctured $3$-torus. (It does retract to a cylinder though, so maybe I could take advantage of that?) I need some way of introducing a boundary on the $3$-torus, just as I did with the $3$-sphere, hopefully in a way that easily generalizes.
Depending on what I find for $3$-dimensional complexes, I may not need to worry about $4$-cells and higher, because I may be able to generalize the construction. For example, I can "push" a $3$-face of a $4$-cell against its other faces in order to produce four $3$-cells, thus reducing the dimension. This operation would fix the $2$-skeleton (and hence $1$-skeleton), just as the above operation fixes the $1$-skeleton. In general, I might be able to find an $f : X \rightarrow \text{skel}^n(X)$ that fixes $\text{skel}^{n-1}(X)$. I would be able to work inductively downwards in this way.
Any advice? Am I approaching this from the completely wrong direction?
(Original question: I have a continuous function $f : \text{skel}^2(X) \rightarrow Y$, where $X$ is a finite simplicial complex and $Y$ is a $2$-dimensional simplicial complex. Can I find a continuous $f' : X \rightarrow Y$ so that $f' = f$ on $\text{skel}^1(X)$? I don't care what $f'$ is on the rest of $\text{skel}^2(X)$. Can I somehow use the fact that $\text{skel}^2(X)$ and $X$ share the same fundamental group, if I can find such an $f'$? Any suggestions on how to approach this?
I already know this cannot be done if we want an $f'$ that agrees with $f$ on all of $\text{skel}^2(X)$, since $f : S^2 \rightarrow S^2$ cannot be extended to $f' : D^3 \rightarrow S^2$. But I am asking for something weaker; I only want and $f'$ that agrees with $f$ on the $1$-skeleton.)