Q:

Is the volume of relative cycles at least the systole of the manifold?

Let $M$ be a manifold with boundary $\partial M$. Suppose that $M$ is equipped with some structure for which a notion of volume for chains can be defined. For example, if $M$ is triangulated, then the volume of a simplicial chain is just the number of cells with non-zero coefficients in the formal sum (let's work over the field of 2 elements for simplicity). An analogous notion can be defined if $M$ has a Riemannian metric instead.
Let $p:C_i(M)\rightarrow C_i(M,\partial M)$ be the canonical map from  $i$-chains to relative $i$-chains, and let $p_*:H_i(M)\rightarrow H_i(M,\partial M) $ the induced map. Define for any $c\in C_i
(M,\partial M)$:
$$vol(c):=\inf\{vol(c') \mid p(c')=c\}$$
Clearly, $p$ cannot increase volume but may very well decrease it. However:

Is it true that the volume of any non-contractible relative $i$-cycle representing an homology class in $Im(p_*)$ is at least $sys_i(M)$?

Here $sys_i(M)$, namely the $i$-systole of $M$, is by definition the infimal volume of a non-contractible $i$-cycle in $M$.
Diagram chasing at the level of chains (and not only at the level of homology) was the main type of arguments I tried to produce for proving such a claim. But I also suspect there might be some not-so-hard counter-example that I'm missing. 

A:

Here is a counterexample. Let $T$ be the $2$-torus obtained by identifying the opposite sides of the square on $\mathbb R^2$ with vertices $(\pm 1, \pm 1)$. Let $B_{1-\varepsilon}$ be the disk $x^2+y^2<1-\varepsilon$ and let $M$ be the complement in $T$, $M=T/B_{1-\varepsilon}$. It is clear that the systole $sys_1(M)$ is equal to $2$. Indeed, this is the length of the shortest non-contractible cycle in $M$. However, if we take such a cycle, for example given by the left side of the square, its image in $H_1(M, \partial M)$ equals the class of the segment of length $2\varepsilon$ that joins two points $(0, -1+\varepsilon)$ and $(0, 1-\varepsilon$) on $M$. Clearly $2\varepsilon<2=sys_1(M)$.

