 Solution

Choose a connected proper incompressible surface $S$ with nonempty boundary as the first cut in a three-manifold hierarchy of $N$. Homotope $f$ relative to $\partial M$ into transverse position and let $F=f^{-1}(S)$.

We first make $F$ incompressible. If $D$ is a compression disk for $F$, its boundary maps to a loop in $S$ which is null-homotopic in $N$. Since $S$ is fundamental-group-injective, this loop is null-homotopic in $S$. Fill it in $S$ and alter $f$ in a small neighbourhood of $D$ so that its inverse image is obtained by compressing $F$ along $D$. The alteration is a homotopy: the filling and the old disk give a sphere map in $N$, which is null-homotopic by the asphericity proved above. A product collar of $S$ specifies the two sides of the modification. The integer ${\sum }_j\max (0,1-\chi (F_j))$ decreases at every essential compression; sphere components can be removed separately. A resulting sphere component of $F$ bounds a three-ball in $M$ by three-manifold irreducibility; move the map of that three-ball to one side of $S$ to remove it. The required extension and homotopy exist because the higher homotopy groups of $N$ vanish. Continuing leaves an incompressible $F$ with no spherical components.

For each component $F_j$, the map on fundamental groups to $S$ is injective. Indeed, its composition with the injection ${\pi }_1(S)\to {\pi }_1(N)$ equals the composition of ${\pi }_1(F_j)\to {\pi }_1(M)$ with the isomorphism $f_{\ast }$. Both maps in the latter composition are injective. A closed nonspherical orientable $F_j$ cannot occur: its fundamental group would inject into ${\pi }_1(S)$, a free group, whereas every subgroup of a free group is free and a closed positive-genus surface group is not free. Thus every component of $F$ has boundary.

Apply the permitted surface-map homotopy result to $F_j\to S$, fixing its boundary. A fundamental-group-injective map of compact surfaces is homotopic either to a covering or, in the annulus case here, to a map into one target boundary circle. The exceptional annulus case is impossible: its two distinct boundary circles would map onto the same circle of $\partial S$, contradicting the injectivity of the original boundary map. If $S$ is a disk, injectivity makes each $F_j$ a disk, and the unique boundary circle directly gives a single disk homeomorphism. Otherwise each component is homotopic to a cover of $S$.

Every point of $\partial S$ has exactly one preimage on $\partial F$, since $f{|}_{\partial M}$ is a homeomorphism. Every cover component contributes its positive degree to the number of preimages on every target boundary circle. Thus there is exactly one component and its covering degree is one. We have proved that $F\to S$ can be homotoped to a homeomorphism, relative to its boundary.

Extend this surface homotopy to $M$, and use the two-sided collars to make $f$ a product homeomorphism on a neighbourhood of $F$. The transverse normal direction is a degree-one local crossing, so the collars on its two sides match the two sides of $S$; any superfluous folds can be removed in the normal interval. We may now cut $M$ along $F$ and $N$ along $S$, obtaining a map whose restriction to the entire cut boundary is a homeomorphism. The other hypotheses also survive: the cut manifolds are irreducible, have nonempty boundary, and the inclusion of every cut component into its uncut manifold is fundamental-group-injective by the normal form of the van Kampen theorem. A loop killed by the cut map is consequently killed by $f_{\ast }$ and hence is trivial in its source cut component.