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It's well known that if $X$ is a CW-complex then $H_n(X;\mathbb{Z})$ is generated at most by $k$-elements, where $k$ is the number of $n$-cells in $X$. Now, I have 2 questions...

$(1)$ Is it true that a generator of $H_n(X)$ cannot be contained in the $n-1$ skeleton of X? If it is true, how I can prove this?

$(2)$ If I have the space $S^n\times S^n$ (CW-complex structure with one $0$-cell , two $n-$cells and one $2n-$cell) and I identify the two $S^n$ (union of a $n-$cell and a $0-$cell) in the CW-complex structure, then I obtain a quentient CW-complex $X$, with projection map $q$ which is a homeomorphism restricted to the $2n$-cell and moreover it maps the two $S^n$ of the product space homeomorphically in a $S^n$ in the quotient. Now if the first question is true, both the generator of $H_n(S^n\times S^n)$ are mapped in the same generator of $H_n(X)$ by $q_\ast$. Is it true the same for the unique generator of $H_{2n}(S^n\times S^n)$, i.e. does $q_\ast$ maps a generator of $H_{2n}(S^n\times S^n)$ onto a generator of $H_{2n}(X)$?

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    What does it mean for a generator of an homology group of a space to be contained in a skeleton of the space? – 2017-02-17
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    You talk about "the generator" of $H_n(S^n\times S^n)$, but that group is isomorphic to $\mathbb Z^2$, so it does not have "a generator", much less a unique one. – 2017-02-17
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    Apart from that, have you tried computing the homology of your two spaces and the map induced by $q$ using the cellular complex? Your questions can be answered by that computation. – 2017-02-17
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    To address question (1) as best as I can, it is a fact that the inclusion map $X^{(n-1)} \to X$ induces the trivial homomorphism $H_n(X^{(n-1)}) \to H_n(X)$. Perhaps you might wish to use this fact to conclude that "a generator of $H_n(X)$ cannot be contained in the $n-1$ skeleton", but that would not really capture the full meaning of this fact. A better statement might be "a nontrivial element of $H_n(X)$ cannot be represented by an $n$-cycle supported in $X^{(n-1)}$". – 2017-02-17
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    Thanks. Your remark is simple, but very helpfull! :) – 2017-02-18

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