2 clarified argument

I think I can prove that $diam(\tilde M)\le m\cdot diam(M)$ for any covering. Let $\tilde p,\tilde q\in\tilde M$ and $\tilde\gamma$ be a shortest path from $\tilde p$ to $\tilde q$. Denote by $p,q,\gamma$ their projections to $M$. I want to prove that $L(\gamma)\le m\cdot diam(M)$. Suppose the contrary.

Split $\gamma$ into $m$ arcs $a_1,\dots,a_n$ of equal length: $\gamma=a_1a_2\dots a_m$. , $L(a_i)=L(\gamma)/m>diam(M)$. Let $b_i$ be a shortest path in $M$ connecting the endpoints of $a_i$. Note that $L(b_i)L(b_i)\le diam(M)< L(a_i)$. I want to replace some of the components $a_i$ of the path $\gamma$ by their "shortcuts" $b_i$ so that the lift of the resulting path starting at $\tilde p$ still ends at $\tilde q$. This will show that $\tilde\gamma$ is not a shortest path from $\tilde p$ to $\tilde q$, a contradiction.

To switch from $a_i$ to $b_i$, you left-multiply $\gamma$ by a loop $l_i:=a_1a_2\dots a_{i-1}b_i(a_1a_2\dots a_i)^{-1}$. More precisely, if you replace the arcs $a_{i_1},a_{i_2},\dots,a_{i_k}$, where $i_1< i_2<\dots< i_k$, by their shortcuts, the resulting path is homotopic to the product $l_{i_1}l_{i_2}\dots l_{i_k}\gamma$. So it suffices to find a product $l_{i_1}l_{i_2}\dots l_{i_k}$ whose lift starting from $\tilde p$ closes up in $\tilde M$. Let $H$ denote the subgroup of $\pi_1(M,p)$ consisting of loops whose lifts starting at $\tilde p$ close up. The index of this subgroup is $m$ since its right cosets are in 1-to-1 correspondence with the pre-images of $p$. While left cosets may be different from right cosets, the number of left cosets is the same $m$.

Now consider the following $m+1$ elements of $\pi_1(M,p)$: $s_0=e$, $s_1=l_1$, $s_2=l_1l_2$, $s_3=l_1l_2l_3$, ..., $s_m=l_1l_2\dots l_m$. Two of them, say $s_i$ and $s_j$ where $i< j$, are in the same left coset. Then $s_i^{-1}s_j=l_{i+1}l_{i+2}\dots l_j\in H$ and we are done.

1

I think I can prove that $diam(\tilde M)\le m\cdot diam(M)$ for any covering. Let $\tilde p,\tilde q\in\tilde M$ and $\tilde\gamma$ be a shortest path from $\tilde p$ to $\tilde q$. Denote by $p,q,\gamma$ their projections to $M$. I want to prove that $L(\gamma)\le m\cdot diam(M)$. Suppose the contrary.

Split $\gamma$ into $m$ arcs $a_1,\dots,a_n$ of equal length: $\gamma=a_1a_2\dots a_m$. Let $b_i$ be a shortest path in $M$ connecting the endpoints of $a_i$. Note that $L(b_i)< L(a_i)$. I want to replace some of the components $a_i$ of the path $\gamma$ by their "shortcuts" $b_i$ so that the lift of the resulting path starting at $\tilde p$ still ends at $\tilde q$. This will show that $\tilde\gamma$ is not a shortest path from $\tilde p$ to $\tilde q$, a contradiction.

To switch from $a_i$ to $b_i$, you left-multiply $\gamma$ by a loop $l_i:=a_1a_2\dots a_{i-1}b_i(a_1a_2\dots a_i)^{-1}$. More precisely, if you replace the arcs $a_{i_1},a_{i_2},\dots,a_{i_k}$, where $i_1< i_2<\dots< i_k$, by their shortcuts, the resulting path is homotopic to the product $l_{i_1}l_{i_2}\dots l_{i_k}\gamma$. So it suffices to find a product $l_{i_1}l_{i_2}\dots l_{i_k}$ whose lift starting from $\tilde p$ closes up in $\tilde M$. Let $H$ denote the subgroup of $\pi_1(M,p)$ consisting of loops whose lifts starting at $\tilde p$ close up. The index of this subgroup is $m$ since its right cosets are in 1-to-1 correspondence with the pre-images of $p$. While left cosets may be different from right cosets, the number of left cosets is the same $m$.

Now consider the following $m+1$ elements of $\pi_1(M,p)$: $s_0=e$, $s_1=l_1$, $s_2=l_1l_2$, $s_3=l_1l_2l_3$, ..., $s_m=l_1l_2\dots l_m$. Two of them, say $s_i$ and $s_j$ where $i< j$, are in the same left coset. Then $s_i^{-1}s_j=l_{i+1}l_{i+2}\dots l_j\in H$ and we are done.