Showing the Sum of $n-1$ Tori is a Double Cover of the Sum of $n$ Copies of $\mathbb{RP}^2$ I want to show that the non-orientable surface of genus $n$ has a 2-sheeted cover by an orientable surface of genus $n-1$.  The base cases are easy: $S^2$ covers $\mathbb{RP}^2$ and I worked on a proof that the torus $\mathbb{T}^2$ covers $\mathbb{RP}^2\#\mathbb{RP}^2\approx K$ (Klein bottle).  
Here are some of my constraints:


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*Nothing about the relationship between the Euler characteristic and covering maps has been proved, so I can't use this.  Nor do I know anything about homology yet.  Additionally, I don't yet know how to produce quotient manifolds using the techniques of group actions.  The placement of this question in the text I'm using suggests that I don't need those tools.  I pretty much only have knowledge of the universal covering space of manifolds.

*I proved previously that if $M$ and $N$ are connected $n$-manifolds and $M$ has a $k$-sheeted covering space $H$, then $H\#N\#\ldots \#N$ (with $k$ copies of $N$) is a covering space for $M\#N$.  


Induction using this fact seems like a good approach.  Suppose $R$ is a connected sum of $n$ projective planes, an $T$ is connected sum of $n-1$ tori such that $T$ is a double cover for $R$.  I want to show that $T\#\mathbb{T}^2$ is a double cover for $R\#\mathbb{RP}^2$.  We know that $T\#\mathbb{RP}^2\#\mathbb{RP}^2 \approx T\#K$ is a double cover for $R\#\mathbb{RP}^2$, but I'm not sure where the go from here.  We can't really cover $T\#K$ by $T\#\mathbb{T}^2$ since there are different numbers of sheets whether you're on $K$ or you're on $T$.  Also, even if we could then we would end up with a quadruple cover of $R\#\mathbb{RP}^2$.
Also if anyone wants to explain to me how $\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2$ is a double cover of $\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2$, I'd greatly appreciate it.
This type have been asked before like here and here, but my knowledge of the techniques used in the first link are scant.  It's suggested that I don't need the techniques of orbit spaces or quotient manifolds to prove this claim, and I would like to develop a proof without those techniques.
 A: Let $kM$ denote the connected sum of $k$ copies of $M$. 
As $3\mathbb{RP}^2 = T^2\#\mathbb{RP}^2$, we see that $(2k - 1)\mathbb{RP}^2 = (k-1)T^2\#\mathbb{RP}^2$ and $2k\mathbb{RP}^2 = (k-1)T^2\# 2\mathbb{RP}^2$.
If $n$ is odd, say $n = 2k - 1$, then $n\mathbb{RP}^2 = (2k-1)\mathbb{RP}^2 = (k-1)T^2\#\mathbb{RP}^2$. As $S^2$ double covers $\mathbb{RP}^2$, we see that $2(k - 1)T^2\# S^2 = (2k - 2)T^2 = (n - 1)T^2$ double covers $n\mathbb{RP}^2$ by the second dot point. 
If $n$ is even, say $n = 2k$, then $n\mathbb{RP}^2 = 2k\mathbb{RP}^2 = (k-1)T^2\# 2\mathbb{RP}^2$. As $T^2$ double covers $2\mathbb{RP}^2$, we see that $2(k - 1)T^2\# T^2 = (2k - 1)T^2 = (n - 1)T^2$ double covers $n\mathbb{RP}^2$ by the second dot point.

As for why $4\mathbb{RP}^2$ double covers $3\mathbb{RP}^2$, note that $3\mathbb{RP}^2 = 2\mathbb{RP}^2\#\mathbb{RP}^2$ and $S^2$ double covers $\mathbb{RP}^2$, so $4\mathbb{RP}^2\# S^2 = 4\mathbb{RP}^2$ double covers $2\mathbb{RP}^2\#\mathbb{RP}^2 = 3\mathbb{RP}^2$ by the second dot point.
