Degree-1 map from connected sum $M\#T^n$ to torus $T^n$

Question

Let $M^n$ be a closed oriented smooth $n$-manifold, denote by $M\#T^n$ its connected sum with the $n$-torus. (How) can I get a smooth degree-1 map $f: M\#T^n \rightarrow T^n$? Are any additional assumptions on $M$ needed in order for $f$ to exist?

Answer 1

Never mind, I was being a moron.

Let $i_1: B \rightarrow M$, $i_2: B \rightarrow T^n$ be (smooth) embeddings of the closed $n$-ball $B$ and set $p_j := i_j(0)$ for $j = 1, 2$.

Then, up to diffeomorphisms, the connected sum $M\#T^n$ is the quotient of the disjoint union of $M\setminus \{p_1\}$ and $T^n \setminus \{p_2\}$ where we identify $i_1(tv) \in M\setminus \{p_1\}$ with $i_2((1-t)v) \in T^n \setminus \{p_2\}$ for every $v \in \partial B = S^{n-1}$ and every $t \in (0,1)$. Now we can define $f: M\#T^n \rightarrow T^n$ in the following way:

$$ f(x) := \begin{cases} x, \text{if } x \in T^n \setminus \{p_2\}, \\ p_2, \text{if } x \in M \setminus \operatorname{int}(i_1(B)) \end{cases} $$

Clearly, every $y \in T^n \setminus \{p_2\}$ is a regular value and has $f^{-1}(\{y\}) = \{y\}$, so $f$ has degree 1 (or -1, depending on which orientation on $M\#T^n$ is chosen; in the latter case one can modify $f$ by swapping two $S¹$ factors in $T^n$ so that is has degree 1).