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I am reading a paper of Waldausen's (https://pub.uni-bielefeld.de/publication/1782175) and I am confused about the proof of Theorem 2.1. In it he has a pair of closed connected compact 3-manifolds $M,N$ a map $f : M \to N$ and a Heegaard splitting $N = X \cup Y$. Waldhausen claims that there is a Heegaard splitting $M = V \cup W$ and a map $g : (M,V,W) \to (N,X,Y)$ (so $g$ preserves the Heegaard splittings) and $g$ is homotopic to $f$.

The proof starts with Waldhausen assuming that $X$ is the regular neighborhood of some 1-complex. Then Waldhausen claims that we may assume that $f^{-1}(X)$ is also a regular neighborhood of some 1-complex. Presumably we have homotoped $f$ to be nicer at this stage. What is the explanation for why we can do this?

Let $U$ be a regular neighborhood of a collection of arcs in $M$ such that the closure of the complement of $f^{-1}(X) \cup U$ is a handlebody. Then Waldhausen claims that we can homotope $f$ so that the inverse image of $X$ is $f^{-1}(X) \cup U$ thus completing the proof. How is this second homotopy done?

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    A degree 1 map: I think the claim just is not true in general. As for your 1st question: It is quite clear that each handlebody in a manifold is the regular neighborhood of a 1-subcomplex (after possibly subdividing the triangulation). Just this subcomplex need not be the entire 1-skeleton of a triangulation. Your second question: Use general position argument to homotop your map to make (after a homotopy) the preimage of a 1-subcomplex again 1-dimensional. – 2017-02-17
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    @MoisheCohen I do not think that the degree 1 assumption is used in the proof of what I mentioned. I edited the question - thank you for the correction! What exactly is the general position argument that is used? – 2017-02-17

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